2.4 Polynomials

Q1

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	N17/5/MATHL/HP1/ENG/TZ0/XX/3
Question Text	(a) Consider the polynomial $q(x) = 2x³ - 5x² - kx + 20$ . Given that $q(x)$ has a factor $(x - 4)$ , find the value of $k$ .
Total Mark	3
Correct Answer	17
Explanation	n/a
Mark Scheme	In this case, the factor of the polynomials is already given. $q(4)=0$ So, $2 \times 4³ - 5 \times 4² - 4k + 20 = 0$ Rearrange to give, $k=17$
Question Text	(b) Hence or otherwise, factorise $q(x$ $)$ as a product of linear factors. Select all the factors. (a) $(x - 4)$ (b) $(x + 1)$ (c) $(x-1)$ (d) $(2x+5)$ (e) $(2x+3)$
Total Mark	3
Correct Answer	(a), (c), (d)
Explanation	n/a
Mark Scheme	If we factorize the given polynomial, we get the following result. $q(x) = (x - 4)(x - 1)(2x + 5)$ Hence, the correct options are (a), (c), (d).

Q2

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	M17/5/MATHL/HP1/ENG/TZ1/XX/12
Question Text	Consider the polynomial $P(z)=z^5+5 z^4+14 z^3+22 z^2+17 z+5, z \in C$ . The polynomial can be written in the form $P(z)=(z+1)^3\left(z^2+b z+c\right)$ . Find the sum of the value of $b$ and the value of $c$ .
Total Mark	5
Correct Answer	7
Explanation	n/a
Mark Scheme	Expand the polynomial form to find the value of $b$ and $c$ . $(z+1)^3\left(z^2+b z+c\right)=z^5+5 z^4+14 z^3+22 z^2+17 z+5 \\ \left(z^3+3 z^2+3 z+1\right)\left(z^2+b z+c\right)=z^5+5 z^4+14 z^3+22 z^2+17 z+5 \\ b=2, c=5$ Hence, the sum of $b$ and $c$ is equal to 7 .

Q3

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	N16-TZ0-P1-5(HL)
Question Text	The quadratic equation $x^2-4 k x+(k-1)=0$ has roots $\alpha$ and $\beta$ such that $\alpha^2+\beta^2=16$ . Without solving the equation, the real number $k$ can be equal to $-\frac{7}{8}$ and $n$ . Find the value of $n$ . Without solving the equation, find the possible values of the real number $k$ .
Total Mark	4
Correct Answer	1
Explanation	n/a
Mark Scheme	We can express the sum and product of $\alpha$ and $\beta$ in terms of $k$ . $\alpha+\beta=4 k \\ \alpha \beta=k-1$ Next, we can express $\alpha^2+\beta^2$ in terms of $k$ by using the sum and product of $\alpha$ and $\beta$ . $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=16 k^2-2 k+2=16 \\ 4 k^2-2 k-2=0$ Hence, if we attempt to solve the quadratic equation. $k=1,-\frac{7}{8}$ The other real number $k$ is equal to 1 .

Q4

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	N16-TZ0-P1-10(HL)
Question Text	A given polynomial function is defined as $f(x)=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n$ . The roots of the polynomial equation $f(x)=0$ are consecutive terms of a geometric sequence with a common ratio of $\frac{1}{3}$ and first term 2 . Given that $a_{n-1}=-80$ and $a_n=27$ (a) Find the degree of the polynomial
Total Mark	4
Correct Answer	4
Explanation	n/a
Mark Scheme	The sum of the roots of the polynomial can be expressed using the following formula: $-\frac{a_{n-1}}{a_n}=\frac{80}{27}$ We can express the sum of the roots of the polynomial in different terms by the geometric sequence summation formula. $2\left(\frac{1-\left(\frac{1}{3}\right)^n}{1-\frac{1}{3}}\right)=\frac{80}{27} \\ 1-\left(\frac{1}{3}\right)^n=\frac{80}{81} \\ \left(\frac{1}{3}\right)^n=\frac{1}{81} \\ n=4$
Question Text	(b) The value of $a_0$ can be expressed as $\frac{p}{q}$ . Find the sum of $p$ and $q$ .
Total Mark	2
Correct Answer	35
Explanation	n/a
Mark Scheme	To find the value of $a_0$ we can use the product of the roots of the polynomial. $(-1)^n \frac{a_0}{a_n}=2 \times \frac{2}{3} \times \frac{2}{9} \times \frac{2}{27}$ If we take into account that $a_n=27$ , $a_0=27 \times 2 \times \frac{2}{3} \times \frac{2}{9} \times \frac{2}{27} \\ a_0=\frac{8}{27}$ The sum of $p$ and $q$ is equal to 35 .

Q5

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	M15-TZ1-P1-7(HL)
Question Text	Let $p(x)=3x⁵ + 6x⁴ - 25x³ - 15x² + 81x + 36 , x ∈ R$ . For the polynomial equation $p(x) = 0$ , state (a) (i) The sum of the roots Total Mark: 2 Correct Answer: -2 Explanation: n/a Mark Scheme: The sum of the roots of $p(x)$ can be expressed using the following formula: $-\frac{a_{n-1}}{a}=-\frac{6}{3}=-2$ (ii) The product of the roots Total Mark: 2 Correct Answer: -12 Explanation: n/a Mark Scheme: The product of the roots of $p(x)$ can be calculated using the following formula: $(-1)^n \frac{a_0}{a_n}=-\frac{36}{3}=-12$
Question Text	(b) A new polynomial is defined by $q(x) = p(x + 4)$ . Find the sum of the roots of the equation $q(x) = 0$ .
Total Mark	2
Correct Answer	-22
Explanation	n/a
Mark Scheme	The roots of $q(x)$ are each 4 less than the roots of $p(x)$ since $p(x+4)$ is the same as shifting $p(x$ ) towards the left by 4 units. Hence, the sum of the roots of $q(x)$ is $-2 - 4 × 5 = -22$

Q6

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	M15-TZ2-P1-12(HL)
Question Text	The cubic equation $x³ + px² + qx + c =$ 0 has roots $α,β,γ$ . By expanding $(x - α)(x - β)(x - γ)$ show that ... (a) Which of the following is equal to $α + β + γ$ ? (a) p (b) -p (c) c (d) -c Total Mark : 2 Correct Answer : b Explanation : na Mark Scheme : (a) (b) (c) Given the three roots $\alpha, \beta, \gamma$ , we can expand the polynomial. $\begin{aligned} x^3+p x^2+q x+c & =(x-\alpha)(x-\beta)(x-\gamma) \\ & =\left(x^2-(\alpha+\beta) x+\alpha \beta\right)(x-\gamma) \\ & =x^3-(\alpha+\beta+\gamma) x^2+(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma \end{aligned}$ If we compare the coefficients, (a) $\alpha+\beta+\gamma=-p$ (b) $\alpha \beta+\beta \gamma+\gamma \alpha=q$ (ic) $\alpha \beta \gamma=-c$ Answer: (a) B, (b) A, (c) D (b) Which of the following is equal to $αβ + βγ + γα$ ? (a) q (b) -q (c) c (d) -c Total Mark : 2 Correct Answer : a Explanation : na Mark Scheme : na (c) Which of the following is equal to $c = -αβγ$ ? (a) p (b) -p (c) c (d) -c Total Mark : 2 Correct Answer : d Explanation: na Mark Scheme: na (d) It is now given that $p$ =-18 and $q$ =27 for parts (d) and (d) below. (i) In the case that the three roots $α,β,γ$ form an arithmetic sequence with a positive common difference. Find the value of the second largest root. Total Mark : 2 Correct Answer: 6 Explanation : na Mark Scheme : Using the information that the three roots form an arithmetic sequence, we can express the three roots as below: $α - d, α, α + d$ We can deduce the value of smallest root by using the sum of the three roots $α - d + α + α + d = 3α = 18$ Hence, the value of the second largest root α is equal to 6. Answer: 6 (ii) Hence determine the value of c. Total Mark : 3 Correct Answer : -270 Explanation : na Mark Scheme : na

Q7

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	N14/5/MATHL/HP1/ENG/TZ0/XX/2
Question Text	The quadratic equation $2 x^2-14 x+3=0$ has roots $\alpha$ and $\beta$ . Another quadratic equation $x^2+p x+q=0$ , $p, q \in Z$ has roots $\frac{3}{\alpha}$ and $\frac{3}{\beta}$ . Find the sum of $p$ and $q$ .
Total Mark	5
Correct Answer	-7
Explanation	n/a
Mark Scheme	Consider the fact $\begin{gathered} \frac{3}{\alpha}+\frac{3}{\beta}=-p \\ \left(\frac{3}{\alpha}\right)\left(\frac{3}{\beta}\right)=q \end{gathered}$ By using the sum and product of the roots, we can express the quadratic equation using $\alpha$ and $\beta$ . $x^2-\left(\frac{3}{\alpha}+\frac{3}{\beta}\right) x+\left(\frac{3}{\alpha}\right)\left(\frac{3}{\beta}\right)=0$ To find the value of $q$ , (Using the formulae for the product of roots, $\alpha \beta=\frac{3}{2}$ .) $q=\frac{9}{\alpha \beta}=6$ To find the value of $p$ , (Using the formulae for the sum of the roots, $\alpha+\beta=7$ ) $p=-\left(\frac{3}{\alpha}+\frac{3}{\beta}\right)=-\frac{3(\alpha+\beta)}{\alpha \beta}=-\frac{3 \times 7}{\frac{3}{2}}=-14$ Hence, the sum of $p$ and $q$ is equal to -8 . Answer: -8

Q8

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	M14/5/MATHL/HP1/ENG/TZ1/X/1
Question Text	When the polynomial $x³ + ax + b$ is divided by $(x - 3)$ , the remainder is 32, and when divided by $x + 2$ , it is 7. Find the sum of $a$ and $b$ .
Total Mark	5
Correct Answer	9
Explanation	n/a
Mark Scheme	na

Q9

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	M14/5/MATHL/HP1/ENG/TZ1/X/4
Question Text	The equation $4 x^3+17 x^2+150 x+3=a$ has roots $n_1, n_2$ and $n_3$ . Given that $n_1+n_2+n_3+n_1 n_2 n_3=0$ , find the value of $a$ . [3]
Total Mark	3
Correct Answer	20
Explanation	n/a
Mark Scheme	$\begin{aligned} & n_1+n_2+n_3=\frac{-17}{4} \\ & n_1 n_2 n_3=\frac{a-3}{4} \\ & \frac{-17}{4}+\frac{a-3}{4}=0 \\ & a=20 \end{aligned}$ Answer: 20

Q10

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	M14/5/MATHL/HP1/ENG/TZ2/XX/4
Question Text	The roots of a quadratic equation $2x² + 10x + 1 = 0$ are $α$ and $β$ . Without solving the equation (a) find the value of $α² + β²$ .
Total Mark	3
Correct Answer	24
Explanation	n/a
Mark Scheme	Use the formulae for the sum and product of roots. $\begin{gathered} \alpha+\beta=-5 \\ \alpha \beta=\frac{1}{2} \end{gathered}$ We can express $\alpha^2+\beta^2$ in terms of $\alpha+\beta$ and $\alpha \beta$ , $\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta$ and substitute the values for the sum and product of the roots. $\alpha^2+\beta^2=(-5)^2-2\left(\frac{1}{2}\right)=24$ Answer: 24
Question Text	(b) A quadratic equation with roots $α²$ and $β²$ can be expressed as $Ax² + Bx + C = 0$ . Find the sum of A and B..
Total Mark	4
Correct Answer	-23
Explanation	n/a
Mark Scheme	Express the quadratic equation with roots $\alpha^2$ and $\beta^2$ , $\left(x-\alpha^2\right)\left(x-\beta^2\right)=x^2-\left(\alpha^2+\beta^2\right) x+\alpha^2 \beta^2$ We can find the coefficients by using the values from question (a) $\begin{gathered} x^2-\left(\alpha^2+\beta^2\right) x+(\alpha+\beta)^2=0 \\ x^2-24 x+\left(\frac{1}{2}\right)^2=0 \\ x^2-24 x+\frac{1}{4}^2=0 \end{gathered}$ Hence, the sum of $A$ and $B$ is equal to -23 . Answer: -23

Q11

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	N13-TZ0-P1-1(HL)
Question Text	The cubic polynomial $3x³ + px² + qx - 5$ has a factor $(x + 5)$ and leaves a remainder 40 when divided by $(x + 3)$ . Find the sum of $p$ and $q$ .
Total Mark	5
Correct Answer	26
Explanation	n/a
Mark Scheme	When $(x+5)$ is a factor of the cubic polynomial, $f(-5) = -375 + 25p - 5q - 5 = 0 25p - 5q = 380$ When $(x + 3)$ leaves a remainder of 40, $f(-3) = - 81 + 9p - 3q - 5 = 40 9p - 3q = 126$ If we attempt to solve simultaneously, $p = 17$ and $q = 9$ . Hence, the sum of p and q is equal to 26. Answer: 26

Q12

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	M13-TZ1-P1-11(HL)
Question Text	Let $q(x) = 3x³ - 11x² - 40x - 12$ . Given that $x = 6$ is a zero of $q$ , find the product of all the solutions of $3x³ - 11x² - 40x - 12 = 0$ .
Total Mark	4
Correct Answer	4
Explanation	n/a
Mark Scheme	Factorise $q(x)$ by $(x - 6)$ . $q(x)=(x - 6)(3x² + 7x + 2)$ Find the remaining roots by expressing q(x) in the factored form. $q(x)=(x - 6)(3x + 1)(x + 2)$ The solutions of $q(x)$ are $x = 6 , x = -13 , x = -2$ . Hence, the product of the solutions is equal to 4. Answer: 4

Q13

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	N18/5/MATHL/HP1/ENG/TZ0/XX/8
Question Text	Consider the equation $z^4+a z^3+b z^2+c z+d=0$ , where $a, b, c, d \in R$ and $z \in C$ . Two of the roots of the equation are $\log _3 6$ and $i \sqrt{2}$ and the sum of all the roots is $3+\log _3 2$ . Find the value of $4 a+d+12=0$ .
Total Mark	7
Correct Answer	4
Explanation	n/a
Mark Scheme	$-i \sqrt{2}$ is a root of the equation. Using the sum of roots, we can also deduce that $3+\log _3 2-\log _3 6$ is a root. To find the value of $a$ , we can use the sum of roots. $\begin{aligned} & -a=3+\log _3 2 \\ & a=-3-\log _3 2 \end{aligned}$ To find the value of $d$ , we can use the product of roots. $(-1)^4 d=2\left(\log _3 6\right)(i \sqrt{2})(-i \sqrt{2})=4 \log _3 6$ Therefore, the value of $4 a+d+12$ is equal to $\begin{gathered} 4\left(-3-\log _3 2\right)+4 \log _3 6+12 \\ -12-4 \log _3 2+4 \log _3 3+4 \log _3 2+12=4 \end{gathered}$ Hence, the answer is equal to 4. Answer: 4

Q14

Topic	2.4 Polynomials
Tag	Polynomials; Domain; Range; Functions; Asymptotes; Graph; Linear; Quadratics; Modulus; Discriminant
Source	M18/5/MATHL/HP1/ENG/TZ1/XX/1
Question Text	Let $f(x) = x⁴ + px³ + qx + 4$ where $p, q$ are constants. The remainder when $f(x)$ is divided by $(x + 2)$ is 7, and the remainder when $f(x)$ is divided by $(x - 1)$ is 1. Find the sum of $p$ and $q$ .
Total Mark	5
Correct Answer	15
Explanation	n/a
Mark Scheme	When $(x + 2)$ leaves a remainder of 7, $f(-2) = 16 - 8p - 2q + 4 = 26$ $-8p -2q = 6$ When $(x - 1)$ leaves a remainder of 40, $f(1) = 1 + p + q + 4 = 20$ $p + q = 15$ If we attempt to solve simultaneously, $p = -6$ and $q = 21$ . Hence, the sum of $p$ and $q$ is equal to 15. Answer: 15