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Polynomials

Practice exam-style IB Math AA questions for Polynomials, aligned with the syllabus and grouped by topic.

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Question 1
HL • Paper 1
Easy
Non Calculator
HL • Paper 1
Easy
Non Calculator

The polynomial function pp is defined by p(x)=k(x+2)2(x−1)3p(x)=k(x+2)^2(x-1)^3, where kk is a constant. The graph of y=p(x)y=p(x) has yy-intercept 1212.

A

kk.

[2]
B

the zeros of pp and their multiplicities.

[2]
C

whether the graph crosses or touches the xx-axis at each zero.

[1]
Question 2
HL • Paper 1
Easy
Non Calculator
HL • Paper 1
Easy
Non Calculator

Consider the polynomial p(x)=x3−3x2−4x+12p(x)=x^3-3x^2-4x+12.

A

x−2x-2 is a factor of p(x)p(x).

[2]
B

all the zeros of pp.

[4]
Question 3
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let p(x)=x3−2x2+3x+6p(x)=x^3-2x^2+3x+6.

A

x+1x+1 is a factor of p(x)p(x).

[2]
B

factorize p(x)p(x) over R\mathbb{R}.

[3]
Question 4
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The equation x3+px2+qx−12=0x^3+px^2+qx-12=0 has roots \lpha\lpha, β\beta and γ\gamma. It is given that \lpha+β+γ=5\lpha+\beta+\gamma=5 and that one root is 33.

A

the value of pp.

[2]
B

qq and the other two roots.

[4]

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Question 5
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A monic quartic polynomial with real coefficients has roots 22, −1-1 and 1+2i1+2i.

A

the polynomial in expanded form.

[5]
Question 6
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The equation 2x5−3x4+7x2−8=02x^5-3x^4+7x^2-8=0 has roots α1,α2,α3,α4,α5\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5.

A

α1+α2+α3+α4+α5\alpha_1+\alpha_2+\alpha_3+\alpha_4+\alpha_5.

[2]
B

The roots of another equation are α1+2,α2+2,α3+2,α4+2,α5+2\alpha_1+2,\alpha_2+2,\alpha_3+2,\alpha_4+2,\alpha_5+2. the sum of the roots of this equation.

[2]
C

α1α2α3α4α5\alpha_1\alpha_2\alpha_3\alpha_4\alpha_5.

[1]
Question 7
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let p(x)=x4+mx3+nx2−7x+6p(x)=x^4+mx^3+nx^2-7x+6. It is given that x−1x-1 is a factor of p(x)p(x) and that the remainder when p(x)p(x) is divided by x−2x-2 is 1212.

A

the values of mm and nn.

[4]
Question 8
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The polynomial p(x)=x4+ax3+bx2−8x+12p(x)=x^4+ax^3+bx^2-8x+12 has remainder 66 when divided by x−1x-1, and remainder 3030 when divided by x+2x+2.

A

Use the remainder theorem to form two equations in aa and bb.

[2]
B

the value of aa and the value of bb.

[2]
C

the remainder when p(x)p(x) is divided by (x−1)(x+2)(x-1)(x+2).

[2]

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Question 9
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The polynomial p(x)=x4+px3+qx2+rx+65p(x)=x^4+px^3+qx^2+rx+65 has real coefficients. Two of its roots are 2+i2+i and 3−2i3-2i.

A

the other two roots of p(x)=0p(x)=0.

[1]
B

the values of pp, qq and rr.

[4]
Question 10
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Let P(x)=x3+ax2+bx+cP(x)=x^3+ax^2+bx+c. When P(x)P(x) is divided by x2−3x+2x^2-3x+2, the remainder is 5x−45x-4. When P(x)P(x) is divided by x−3x-3, the remainder is 1111.

A

three equations in aa, bb and cc.

[3]
B

aa, bb and cc.

[2]
C

P(4)P(4).

[1]
Question 11
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A monic quartic polynomial has real coefficients, constant term 2020, and roots 1+2i1+2i, 22 and 22.

A

the fourth root of the polynomial.

[1]
B

the polynomial in expanded form.

[4]
Question 12
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The polynomial p(x)=x3+ax2+bx+7p(x)=x^3+ax^2+bx+7 leaves the same remainder when divided by each of x−1x-1, x−3x-3 and x−5x-5.

A

aa and bb.

[3]
B

the common remainder.

[1]
C

factorize p(x)−22p(x)-22.

[1]

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Question 13
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the polynomial

p(x)=x4−5x3−7x2+41x−30p(x)=x^4-5x^3-7x^2+41x-30
A

Use your GDC to the zeros of pp.

[2]
B

the inequality p(x)≤0p(x)\le 0.

[3]
Question 14
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A quartic polynomial PP has real coefficients and positive leading coefficient. It has a double root at x=1x=1, is divisible by x2+4x^2+4, and satisfies P(0)=12P(0)=12.

A

P(x)P(x) in factorized form.

[3]
B

Write P(x)P(x) in expanded form.

[1]
C

the real zeros of PP and their multiplicities.

[1]
Question 15
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The polynomial p(x)=x4+kx3−3x2+lx+2p(x)=x^4+kx^3-3x^2+lx+2 has (x−1)2(x-1)^2 as a factor.

A

the values of kk and ll.

[5]
Question 16
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The roots of p(x)=x3−2x2−5x+6p(x)=x^3-2x^2-5x+6 are α\alpha, β\beta and γ\gamma.

A

a monic polynomial equation with roots α+1\alpha+1, β+1\beta+1 and γ+1\gamma+1.

[5]

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Question 17
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let p(x)p(x) be the monic cubic polynomial p(x)=x3+ax2+bx+cp(x)=x^3+ax^2+bx+c. The sum of the roots of p(x)=0p(x)=0 is 44. It is also given that x−2x-2 is a factor of p(x)p(x) and that the remainder when p(x)p(x) is divided by x+1x+1 is 66.

A

aa, bb and cc.

[4]
B

the other two roots of p(x)=0p(x)=0.

[2]
Question 18
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

Consider the polynomial

p(x)=x5−4x4+2x3+10x2−27x+18p(x)=x^5-4x^4+2x^3+10x^2-27x+18
A

Use your GDC to all the real zeros of pp.

[2]
B

factorize p(x)p(x) over R\mathbb{R}.

[3]
C

Use the product of the roots to that the factorization is consistent with the constant term of p(x)p(x).

[1]
Question 19
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The graph of y=f(x)y=f(x), where

f(x)=x4−4x3−2x2+12x+5f(x)=x^4-4x^3-2x^2+12x+5

intersects a horizontal line y=cy=c.

A

the coordinates of the stationary points of the graph of y=f(x)y=f(x).

[3]
B

the range of values of cc for which f(x)=cf(x)=c has exactly four distinct real roots.

[1]
C

the values of cc for which f(x)=cf(x)=c has exactly two distinct real roots.

[2]
Question 20
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The cubic polynomial p(x)=x3−3x2+kx+4p(x)=x^3-3x^2+kx+4 has a repeated root.

A

the value of kk.

[3]
B

factorize p(x)p(x) completely over R\mathbb{R}.

[2]

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Question 21
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The polynomial

p(x)=2x5−7x4+ax3+bx2+cx−75p(x)=2x^5-7x^4+ax^3+bx^2+cx-75

has real coefficients. Four of its roots are 2+i2+i, 2−i2-i, −1-1 and 33.

A

the fifth root of p(x)=0p(x)=0.

[3]
B

your answer using the product of the roots.

[2]
Question 22
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A monic quartic polynomial PP has real coefficients and is divisible by x2−2x+5x^2-2x+5. The remainders when P(x)P(x) is divided by x−1x-1 and by x+1x+1 are 1212 and 7272 respectively.

A
I.

Write P(x)P(x) in the form (x2−2x+5)(x2+px+q)(x^2-2x+5)(x^2+px+q). Use the given remainders to form two equations in pp and qq.

[3]
II.

pp and qq.

[1]
B
I.

Factorize P(x)P(x) over R\mathbb{R}.

[2]
II.

all the roots of P(x)=0P(x)=0.

[4]
Question 23
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A polynomial function pp of degree 55 has exactly two distinct zeros. The zero at x=−2x=-2 has multiplicity 22, and the zero at x=1x=1 has multiplicity 33. It is also given that p(0)=16p(0)=16. Define q(x)=p(2x−1)q(x)=p(2x-1).

A
I.

p(x)p(x) in factorized form.

[3]
II.

whether the graph of y=p(x)y=p(x) crosses or touches the xx-axis at each zero.

[1]
B
I.

the zeros of qq and their multiplicities.

[3]
II.

q(0)q(0).

[2]
Question 24
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

A cubic polynomial f(x)=ax3+bx2+cx+df(x)=ax^3+bx^2+cx+d satisfies

f(0)=1,f(1)=1,f(2)=3,f(3)=13f(0)=1,\quad f(1)=1,\quad f(2)=3,\quad f(3)=13
A
I.

Use the given values to form equations for aa, bb, cc and dd.

[2]
II.

f(x)f(x).

[3]
B
I.

f(4)f(4).

[1]
II.

Factorize f(x)−1f(x)-1 completely over R\mathbb{R}, and the roots of f(x)=1f(x)=1.

[3]

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Question 25
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

Let p(x)=x4−3x3+ax2+bx+5p(x)=x^4-3x^3+ax^2+bx+5. The remainder when p(x)p(x) is divided by x−1x-1 is 44, and the remainder when p(x)p(x) is divided by x+2x+2 is 6161.

A

the values of aa and bb.

[4]
B

the remainder when p(x)p(x) is divided by x2+x−2x^2+x-2.

[3]
Question 26
HL • Paper 1
Medium
Non Calculator
HL • Paper 1
Medium
Non Calculator

The equation x3−8x2+kx−12=0x^3-8x^2+kx-12=0 has roots \lpha\lpha, β\beta and γ\gamma. It is given that γ=\lpha+β\gamma=\lpha+\beta.

A

the value of γ\gamma.

[2]
B

kk and the other two roots.

[4]
Question 27
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

A monic quartic polynomial pp has real coefficients. Two of the roots of p(x)=0p(x)=0 are 1+2i1+2i and 22. The sum of all four roots is 77 and the product of all four roots is 3030.

A
I.

the conjugate non-real root.

[1]
II.

the remaining real root.

[3]
B
I.

p(x)p(x) in expanded form.

[3]
II.

Let q(x)=p(x+1)q(x)=p(x+1). the roots of q(x)=0q(x)=0.

[1]
Question 28
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

For real aa, define pa(x)=x4+ax3−6x2+4x+8p_a(x)=x^4+ax^3-6x^2+4x+8.

A
I.

if x=2x=2 is a repeated root of pa(x)=0p_a(x)=0, then a=−1a=-1.

[3]
II.

that x=2x=2 is a repeated root when a=−1a=-1.

[1]
B
I.

Factorize p−1(x)p_{-1}(x) completely over R\mathbb{R}.

[2]
II.

the zeros of p−1p_{-1} and their multiplicities, and the behaviour of the graph at each zero.

[2]

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Question 29
HL • Paper 2
Medium
Calculator Permitted
HL • Paper 2
Medium
Calculator Permitted

The equation

x3+ax2+44x−48=0x^3+ax^2+44x-48=0

has three positive real roots which are in arithmetic progression.

A

Let the roots be m−dm-d, mm and m+dm+d. mm satisfies m3−22m+24=0m^3-22m+24=0.

[3]
B

Use your GDC to the roots of the original equation.

[2]
C

the value of aa.

[1]
Question 30
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The monic cubic polynomial p(x)=x3+ax2+bx−27p(x)=x^3+ax^2+bx-27 has roots 3r\frac{3}{r}, 33 and 3r3r, where r≠0r\ne 0. It is given that the remainder when p(x)p(x) is divided by x−1x-1 is −2-2.

A
I.

Use the product of the roots to that 33 is the middle root in this geometric sequence.

[2]
II.

r+1r=3r+\frac{1}{r}=3.

[3]
B
I.

the values of aa and bb.

[3]
II.

the three roots of p(x)=0p(x)=0.

[4]
Question 31
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let P(x)=x4+ax3+bx2+cx+10P(x)=x^4+ax^3+bx^2+cx+10, where a,b,c∈Ra,b,c\in\mathbb{R}. It is given that x2+1x^2+1 is a factor of P(x)P(x) and that the remainder when P(x)P(x) is divided by x−1x-1 is 1010.

A
I.

Use the factor theorem to obtain two equations involving aa, bb and cc.

[3]
II.

aa, bb and cc.

[3]
B
I.

the remainder when P(x)P(x) is divided by x2−3x+2x^2-3x+2.

[3]
II.

x2−3x+2x^2-3x+2 is a factor of P(x)−10P(x)-10.

[1]
Question 32
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let p(x)=x3−5x2+2x+8p(x)=x^3-5x^2+2x+8. The roots of p(x)=0p(x)=0 are denoted by α\alpha, β\beta and γ\gamma. A new polynomial equation has roots 1α−1\frac{1}{\alpha-1}, 1β−1\frac{1}{\beta-1} and 1γ−1\frac{1}{\gamma-1}.

A
I.

x−4x-4 is a factor of p(x)p(x).

[2]
II.

factorize p(x)p(x) completely over R\mathbb{R}.

[3]
B
I.

the three roots of the new equation.

[2]
II.

a polynomial equation with integer coefficients having these roots.

[4]

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Question 33
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let

p(x)=x5−2x4+ax3+bx2+cx+12p(x)=x^5-2x^4+ax^3+bx^2+cx+12

where a,b,c∈Ra,b,c\in\mathbb{R}. It is given that x2−4x^2-4 is a factor of p(x)p(x) and that the remainder when p(x)p(x) is divided by x+1x+1 is 3030.

A
I.

Use the factor theorem to form three equations in aa, bb and cc.

[3]
II.

these equations to aa, bb and cc.

[3]
B
I.

factorize p(x)p(x) over R\mathbb{R}.

[4]
II.

all the roots of p(x)=0p(x)=0.

[3]
Question 34
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

The monic cubic polynomial p(x)=x3+ax2+bx−80p(x)=x^3+ax^2+bx-80 has three roots which are consecutive terms of an arithmetic sequence. It is given that the sum of the roots is 1515.

A
I.

aa.

[2]
II.

Let the roots be m−dm-d, mm and m+dm+d. the three roots of p(x)=0p(x)=0.

[4]
B
I.

bb.

[2]
II.

A second monic cubic equation has roots equal to the squares of the roots of p(x)=0p(x)=0. this equation.

[4]
Question 35
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let p(x)=x3−4x2+x+6p(x)=x^3-4x^2+x+6. The roots of p(x)=0p(x)=0 are α\alpha, β\beta and γ\gamma. A polynomial equation q(x)=0q(x)=0 has roots 1α\frac{1}{\alpha}, 1β\frac{1}{\beta} and 1γ\frac{1}{\gamma}.

A
I.

x+1x+1 is a factor of p(x)p(x).

[2]
II.

Factorize p(x)p(x) completely over R\mathbb{R}.

[2]
B
I.

the roots of q(x)=0q(x)=0.

[2]
II.

a polynomial equation with integer coefficients for q(x)=0q(x)=0.

[4]
Question 36
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

A monic quartic polynomial PP has real coefficients. Two of its roots are 2+i2+i and 2−i2-i. The other two roots are equal. The coefficient of x3x^3 in P(x)P(x) is −8-8.

A
AI.

Let the repeated root be rr. r=2r=2.

[3]
AII.

Write P(x)P(x) in factorized form over R\mathbb{R}.

[1]
B
BI.

P(x)P(x) in expanded form.

[3]
BII.

the remainder when P(x)P(x) is divided by x+1x+1.

[2]
BIII.

the product of all four roots of P(x)=0P(x)=0 and that it is consistent with the expanded polynomial.

[1]

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Question 37
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Consider the polynomial p(x)=x5+ax4+bx3−7x2+cx+10p(x)=x^5+ax^4+bx^3-7x^2+cx+10, where aa, bb and cc are real constants. It is given that x2−1x^2-1 is a factor of p(x)p(x) and that the remainder when p(x)p(x) is divided by x−2x-2 is 1212.

A
I.

Use the factor theorem and the remainder theorem to form three equations in aa, bb and cc.

[3]
II.

aa, bb and cc.

[2]
B
I.

write p(x)p(x) as the product of x2−1x^2-1 and a cubic polynomial.

[2]
II.

Use your GDC to all real zeros of pp, giving your answers to three significant figures.

[2]
III.

the number of non-real roots of p(x)=0p(x)=0.

[1]
Question 38
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The polynomial p(x)=x4−15x3+70x2−120x+64p(x)=x^4-15x^3+70x^2-120x+64 has roots α1\alpha_1, α2\alpha_2, α3\alpha_3 and α4\alpha_4. A new polynomial qq has roots βi=1+1αi\beta_i=1+\dfrac{1}{\alpha_i}, for i=1,2,3,4i=1,2,3,4.

A
I.

Use your GDC or otherwise to the roots of p(x)=0p(x)=0.

[1]
II.

the four roots of q(x)=0q(x)=0.

[2]
B
I.

the sum and product of the roots of q(x)=0q(x)=0.

[2]
II.

the monic polynomial q(x)q(x) in expanded form.

[3]
Question 39
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A cubic polynomial has leading coefficient 22. When divided by x−3x-3 and by x+2x+2, the remainder is 00 in each case. When divided by x−1x-1, the remainder is 44.

A
I.

Write the polynomial in the form p(x)=2(x−3)(x+2)(x−r)p(x)=2(x-3)(x+2)(x-r) and rr.

[3]
II.

write p(x)p(x) in expanded form.

[1]
B
I.

Use your graphing calculator to p(x)>0p(x)>0.

[3]
II.

the sum of the roots of p(x)=0p(x)=0 and it using the coefficient of x2x^2.

[1]
Question 40
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A symmetric bridge arch is modelled by the polynomial h(x)=ax4+bx2+ch(x)=ax^4+bx^2+c, where hh is the height in metres and xx is the horizontal distance in metres from the centre of the arch. The arch is 66 metres high at its centre. At x=4x=4, the arch meets the ground and has gradient −5-5.

A
I.

the value of cc.

[1]
II.

Use the information at x=4x=4 to aa and bb.

[3]
B
I.

the width of the arch at a height of 33 metres.

[3]
II.

the mathematical reason why the two values of xx found in part (b)(i) are opposite in sign.

[1]

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Question 41
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let f(x)=x4−4x2+1f(x)=x^4-4x^2+1. The graph of y=f(x)y=f(x) is intersected by the horizontal line y=cy=c. The graph includes an illustrative horizontal line at y=−1y=-1 (that is, c=−1c=-1); it is not intended to represent a general value of cc.

A
I.

the coordinates of the stationary points of y=f(x)y=f(x).

[3]
II.

Classify the stationary point at x=0x=0.

[1]
B
I.

the range of values of cc for which f(x)=cf(x)=c has four distinct real roots.

[2]
II.

the values of cc for which f(x)=cf(x)=c has exactly two distinct real roots.

[2]
Question 42
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

A monic polynomial PP of degree 55 has real coefficients. It has a double root at x=1x=1, a root at x=−3x=-3, and a non-real root 1+3i1+3i.

A
I.

the remaining non-real root.

[1]
II.

Write P(x)P(x) in factorized form over R\mathbb{R}.

[3]
B
I.

Expand P(x)P(x).

[2]
II.

Use Vieta's formulae to the sum and product of the five roots.

[2]
Question 43
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The diagram shows the family of quartic polynomial functions

ft(x)=x4−2tx2+1,t∈Rf_t(x)=x^4-2tx^2+1, \qquad t\in \mathbb{R}

The value of tt changes the number and nature of the real zeros of the graph.

A
I.

the equation ft(x)=0f_t(x)=0 may be written as a quadratic equation in uu, where u=x2u=x^2.

[2]
II.

the roots of this quadratic equation in terms of tt.

[3]
B

the values of tt for which ft(x)=0f_t(x)=0 has four distinct real roots.

[3]
C
I.

all repeated real roots of ft(x)=0f_t(x)=0 and the corresponding values of tt.

[3]
II.

the multiplicity of each real root when t=1t=1.

[1]
D

For t>1t>1, let the four real roots be −b-b, −a-a, aa and bb, where 0<a<b0<a<b. ab=1ab=1, and a+ba+b in terms of tt.

[2]
Question 44
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A monic cubic polynomial PP is to be constructed from information about remainders. It is given that

P(x)=x3+ax2+bx+cP(x)=x^3+ax^2+bx+c

The graph of y=P(x)y=P(x) is shown together with the horizontal line y=4y=4.

A
I.

The remainder when P(x)P(x) is divided by x−1x-1 is 44. the corresponding equation in aa, bb and cc.

[1]
II.

The remainder when P(x)P(x) is divided by x+2x+2 is −18-18. the corresponding equation in aa, bb and cc.

[1]
III.

It is also given that P(0)=2P(0)=2. aa, bb and cc.

[2]
B

Let Q(x)=3P(x)−12Q(x)=3P(x)-12.

Q(x)=3x3−5x2+8x−6Q(x)=3x^3-5x^2+8x-6
[2]
C
I.

Use your GDC to the real root of Q(x)=0Q(x)=0.

[2]
II.

what this root represents on the graph of y=P(x)y=P(x).

[2]
D

A second monic cubic RR satisfies R(x)−4=(x−s)2(x−t)R(x)-4=(x-s)^2(x-t), where s≠ts\ne t. Given that R(1)=4R(1)=4 and R(−2)=−14R(-2)=-14, all possible pairs (s,t)(s,t).

[3]

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Question 45
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A polynomial equation has roots α1,α2,…,αn\alpha_1,\alpha_2,\ldots,\alpha_n. A new equation is formed by translating each root 33 units to the right. This question investigates how the coefficients change under this transformation.

A
I.

The monic cubic p(x)=x3−6x2+11x−6p(x)=x^3-6x^2+11x-6 has roots α\alpha, β\beta and γ\gamma. α+β+γ\alpha+\beta+\gamma and αβγ\alpha\beta\gamma.

[2]
II.

the monic cubic whose roots are α+3\alpha+3, β+3\beta+3 and γ+3\gamma+3.

[2]
B

A monic polynomial p(x)p(x) of degree nn has roots α1,α2,…,αn\alpha_1,\alpha_2,\ldots,\alpha_n and coefficient of xn−1x^{n-1} equal to AA. the coefficient of xn−1x^{n-1} in the monic polynomial whose roots are α1+3,α2+3,…,αn+3\alpha_1+3,\alpha_2+3,\ldots,\alpha_n+3.

[3]
C
I.

Let p(x)=x4−4x3+kx2+mx+9p(x)=x^4-4x^3+kx^2+mx+9. Given that the roots of p(x)=0p(x)=0 are translated 33 units to the right to form q(x)q(x), the coefficient of x3x^3 in q(x)q(x).

[2]
II.

If the product of the roots of q(x)=0q(x)=0 is 00, one root of p(x)=0p(x)=0 is −3-3.

[3]
Question 46
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A polynomial PP is divided by a quadratic divisor. The quotient and remainder are represented in the diagram by the identity

P(x)=(x2−4)Q(x)+R(x)P(x)=(x^2-4)Q(x)+R(x)

where the degree of RR is less than 22.

A
I.

why R(x)R(x) may be written in the form mx+nmx+n.

[1]
II.

It is given that P(2)=9P(2)=9 and P(−2)=−7P(-2)=-7. R(x)R(x).

[3]
B

Let

P(x)=x4+ax3+bx2+cx+dP(x)=x^4+ax^3+bx^2+cx+d

It is also given that x−1x-1 and x+3x+3 are factors of P(x)P(x), and that the remainder on division by x2−4x^2-4 is the function found in part (a). aa, bb, cc and dd.

[5]
C
I.

factorize P(x)P(x) over R\mathbb{R}.

[2]
II.

the zeros of PP.

[1]
Question 47
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The points (0,P(0))(0,P(0)), (1,P(1))(1,P(1)) and (2,P(2))(2,P(2)) lie on a straight line. It is known that P(0)=5P(0)=5, P(1)=9P(1)=9 and P(2)=13P(2)=13.

A
I.

the linear polynomial R(x)R(x) passing through the three points (0,5)(0,5), (1,9)(1,9) and (2,13)(2,13).

[2]
II.

why P(x)−R(x)P(x)-R(x) has factors xx, x−1x-1 and x−2x-2.

[2]
B

Assume PP is a monic cubic. P(x)P(x).

[3]
C
I.

Use your GDC to the real zero of PP.

[2]
II.

why the other two roots are non-real.

[1]
D

the remainder when P(x)P(x) is divided by x2−3x+2x^2-3x+2.

[2]
Question 48
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Consider the reciprocal polynomial equation

2x4−9x3+14x2−9x+2=02x^4-9x^3+14x^2-9x+2=0
A
AI.

x=0x=0 is not a root, and divide the equation by x2x^2.

[2]
AII.

Let u=x+1xu=x+\frac{1}{x}. uu satisfies 2u2−9u+10=02u^2-9u+10=0.

[3]
B
BI.

the quadratic equation for uu.

[2]
BII.

the original quartic equation.

[3]

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Question 49
HL • Paper 1
Hard
Non Calculator
HL • Paper 1
Hard
Non Calculator

Let p(x)=x4+ax3+bx2−2x+3p(x)=x^4+ax^3+bx^2-2x+3, where a,b∈Ra,b\in\mathbb{R}. The roots of p(x)=0p(x)=0 are α1,α2,α3,α4\alpha_1,\alpha_2,\alpha_3,\alpha_4.

A
I.

the roots of p(x−1)=0p(x-1)=0 are α1+1,α2+1,α3+1,α4+1\alpha_1+1,\alpha_2+1,\alpha_3+1,\alpha_4+1.

[2]
II.

Given that the sum of the roots of p(x−1)=0p(x-1)=0 is 1010, aa.

[2]
B
I.

the product of the roots of p(x+1)=0p(x+1)=0 is p(1)p(1).

[2]
II.

Given that this product is 1212, bb.

[3]
Question 50
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let P(x)=x4+ax3+bx2+cx+dP(x)=x^4+ax^3+bx^2+cx+d. When P(x)P(x) is divided by x2−1x^2-1, the remainder is 3x+43x+4. When P(x)P(x) is divided by x2+1x^2+1, the remainder is 2x−52x-5.

A
I.

Use the first division statement to obtain two equations in aa, bb, cc and dd.

[2]
II.

Use the second division statement to obtain two further equations.

[2]
III.

aa, bb, cc and dd.

[1]
B
I.

Write P(x)P(x) explicitly.

[1]
II.

Use your GDC to the real zeros of PP, giving your answers to three significant figures.

[2]
III.

the number of non-real roots of P(x)=0P(x)=0.

[1]
Question 51
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

The equation x3−14x2+kx−64=0x^3-14x^2+kx-64=0 has three positive real roots which form a geometric progression. Let the roots be ar\dfrac{a}{r}, aa and arar, where a>0a>0 and r>1r>1.

A
I.

a=4a=4.

[2]
II.

2r2−5r+2=02r^2-5r+2=0.

[2]
B
I.

the three roots of the cubic equation.

[2]
II.

the value of kk.

[2]
Question 52
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

For real aa, consider the cubic polynomial fa(x)=x3−3x+af_a(x)=x^3-3x+a.

A
I.

the coordinates of the stationary points of y=fa(x)y=f_a(x) in terms of aa.

[3]
II.

which stationary point is a local maximum.

[1]
B
I.

the range of values of aa for which fa(x)=0f_a(x)=0 has three distinct real roots.

[2]
II.

For a=1a=1, use your GDC to the three roots of fa(x)=0f_a(x)=0.

[2]

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Question 53
HL • Paper 2
Hard
Calculator Permitted
HL • Paper 2
Hard
Calculator Permitted

Let p(x)=x5−3x4+2x3+5x2−7x+4p(x)=x^5-3x^4+2x^3+5x^2-7x+4. The roots of p(x)=0p(x)=0 are α1,α2,α3,α4,α5\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5. For x≠0x\ne0, define q(x)=x5p(1x)q(x)=x^5p\left(\dfrac{1}{x}\right); let qq be the polynomial extension of this expression to x=0x=0.

A

Part (a)

I.

q(x)q(x) in expanded form.

[2]
II.

that the roots of q(x)=0q(x)=0 are 1α1,1α2,1α3,1α4,1α5\dfrac{1}{\alpha_1},\dfrac{1}{\alpha_2},\dfrac{1}{\alpha_3},\dfrac{1}{\alpha_4},\dfrac{1}{\alpha_5}.

[2]
B

Part (b)

I.

∑i=151αi\displaystyle\sum_{i=1}^{5}\dfrac{1}{\alpha_i}.

[2]
II.

∏i=151αi\displaystyle\prod_{i=1}^{5}\dfrac{1}{\alpha_i}.

[2]
Question 54
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A reciprocal quartic is a quartic polynomial whose coefficients read the same forwards and backwards. Consider

P(x)=x4+ax3+bx2+ax+1P(x)=x^4+ax^3+bx^2+ax+1

where a,b∈Ra,b\in \mathbb{R}. The graph shown is for one choice of aa and bb.

A
I.

For x≠0x\ne 0, P(x)=0P(x)=0 is equivalent to

x2+ax+b+a1x+1x2=0x^2+ax+b+a\frac{1}{x}+\frac{1}{x^2}=0
[2]
II.

Let u=x+1xu=x+\frac{1}{x}. P(x)=0P(x)=0 becomes

u2+au+(b−2)=0u^2+au+(b-2)=0
[3]
B

For a=−5a=-5 and b=7b=7, all roots of P(x)=0P(x)=0 exactly.

[5]
C
I.

why a real value of u=x+1xu=x+\frac{1}{x} can produce real values of xx only when u≤−2u\le -2 or u≥2u\ge 2.

[2]
II.

the number of real roots of P(x)=0P(x)=0 when the two roots of u2+au+(b−2)=0u^2+au+(b-2)=0 are 33 and −1-1.

[1]
D

Suppose P(x)=0P(x)=0 has roots rr, 1r\frac{1}{r}, ss and 1s\frac{1}{s}, where r,s≠0r,s\ne 0. Use the product of the roots to why the constant term of P(x)P(x) is 11.

[2]
Question 55
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A sensor response is modelled by a cubic polynomial S(x)S(x) on the interval 0≤x≤60\le x\le 6. The graph appears to touch the xx-axis at x=2x=2 and cross it at x=5x=5. A data table gives the additional value S(0)=20S(0)=20.

A
I.

why S(x)S(x) can be written in the form k(x−2)2(x−5)k(x-2)^2(x-5).

[2]
II.

Use S(0)=20S(0)=20 to kk.

[2]
B

S(x)S(x) in expanded form.

[2]
C
I.

A horizontal threshold line y=hy=h is used. Use your GDC to the local maximum value of S(x)S(x) on 0≤x≤60\le x\le 6.

[2]
II.

the number of solutions of S(x)=hS(x)=h in 0≤x≤60\le x\le 6 when 0<h<40<h<4.

[2]
D

The model is modified to T(x)=S(x)+mT(x)=S(x)+m. the range of mm for which T(x)=0T(x)=0 has exactly two distinct real roots.

[3]
Question 56
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A monic quintic polynomial PP has real coefficients. Two of its roots are 1+2i1+2i and 3−i3-i. The remaining real root is denoted by rr.

AI.

two other non-real roots of P(x)=0P(x)=0.

[1]
AII.

the product of the four non-real roots.

[2]
B

The constant term of P(x)P(x) is −400-400. rr.

[3]
CI.

P(x)P(x) in factorized form over R\mathbb{R}.

[3]
CII.

the coefficient of x4x^4 in P(x)P(x).

[2]
D

Another monic quintic with the same four non-real roots has coefficient of x4x^4 equal to −13-13. its constant term.

[3]

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Question 57
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

The roots of a monic cubic polynomial are positive and in geometric progression. Let the roots be mr\frac{m}{r}, mm and mrmr, where m>0m>0 and r>1r>1, so that they are in increasing order as shown.

A
I.

the product of the three roots is m3m^3.

[1]
II.

The polynomial is x3−Ax2+Bx−64x^3-Ax^2+Bx-64. mm.

[2]
III.

Express AA and BB in terms of rr.

[2]
B

B=4AB=4A for any such cubic.

[2]
C
I.

Consider x3−21x2+84x−64=0x^3-21x^2+84x-64=0. that its positive roots are in geometric progression.

[2]
II.

the common ratio of the progression.

[2]
D

Use your GDC to whether x3−18x2+72x−64=0x^3-18x^2+72x-64=0 has three positive roots in geometric progression. your answer.

[3]
Question 58
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A sequence of values is generated by a polynomial function p(n)p(n), where nn is a positive integer. The table shows values of p(n)p(n) and successive differences.

nn

p(n)p(n)

First diff

Second diff

Third diff

1

0

6

12

6

2

6

18

18

6

3

24

36

24

4

60

60

5

120

A
I.

The values p(1)=0p(1)=0, p(2)=6p(2)=6, p(3)=24p(3)=24 and p(4)=60p(4)=60 are given. why these values are consistent with a cubic polynomial whose constant third difference is 66.

[2]
II.

the leading coefficient of this cubic polynomial.

[2]
B

Let p(x)=x3+ax2+bx+cp(x)=x^3+ax^2+bx+c. Use the given values to p(x)p(x).

[4]
C
I.

Factorize p(x)p(x) completely over R\mathbb{R}.

[1]
II.

all integer roots of p(x)=0p(x)=0.

[2]
D

A quartic polynomial qq has the same first four values as pp, but also q(5)=130q(5)=130. why q(x)−p(x)q(x)-p(x) must have factors x−1x-1, x−2x-2, x−3x-3 and x−4x-4, and a possible expression for q(x)q(x).

[2]
Question 59
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A non-zero root transformation is defined as follows. If p(x)=anxn+an−1xn−1+⋯+a0p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 has non-zero roots α1,α2,…,αn\alpha_1,\alpha_2,\ldots,\alpha_n, then a new polynomial is formed whose roots are 1α1,1α2,…,1αn\frac{1}{\alpha_1},\frac{1}{\alpha_2},\ldots,\frac{1}{\alpha_n}.

A
I.

a polynomial with roots 1α1,1α2,…,1αn\frac{1}{\alpha_1},\frac{1}{\alpha_2},\ldots,\frac{1}{\alpha_n} is proportional to xnp(1x)x^n p\left(\frac{1}{x}\right).

[3]
II.

Write xnp(1x)x^n p\left(\frac{1}{x}\right) in terms of the coefficients of pp.

[2]
B

Let p(x)=2x3−7x2+5x−3p(x)=2x^3-7x^2+5x-3. a cubic polynomial with integer coefficients whose roots are the reciprocals of the roots of p(x)=0p(x)=0.

[3]
C
I.

If the roots of p(x)=0p(x)=0 are α\alpha, β\beta and γ\gamma, 1α+1β+1γ\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} without solving the cubic.

[2]
II.

the product of the reciprocal roots.

[2]
D

A monic polynomial has non-zero roots and is unchanged by replacing every root by its reciprocal. a condition on its constant term.

[2]
Question 60
HL • Paper 3
Hard
Calculator Permitted
HL • Paper 3
Hard
Calculator Permitted

A monic quartic polynomial PP is required to touch the xx-axis at two distinct points. Let these points be x=ax=a and x=bx=b, where a≠ba\ne b. The graph shown is illustrative only and corresponds to the possible branch a=1a=1, b=5b=5; it does not imply that this is the only possible pair.

A
I.

why P(x)P(x) must have the form (x−a)2(x−b)2(x-a)^2(x-b)^2.

[2]
II.

Expand P(x)P(x) in terms of s=a+bs=a+b and p=abp=ab.

[2]
B

A particular polynomial of this type has coefficient of x3x^3 equal to −12-12 and constant term 2525. all possible values of aa and bb.

[4]
C
I.

For each polynomial obtained in part (b), P(x)P(x) in expanded form.

[2]
II.

For either polynomial obtained in part (b), use your GDC to the xx-coordinate of the local maximum between the two touching points.

[2]
D

that, for any distinct real aa and bb, the local maximum between the two touching points occurs at x=a+b2x=\frac{a+b}{2}.

[3]

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Modulus & Inequalities

Properties of Functions