CAIE IGCSE Additional Math 3 factors of polynomials

Use this Polynomial Factors hub to move from remainder and factor theorem conditions into quadratic factors and cubic-equation solutions.

Covered syllabus points

3.1. Use remainder and factor theorems 3.2. Find polynomial factors 3.3. Solve cubic equations

[Maximum number: 5]

The polynomial p is given by $\mathrm{p}(x)=a^{2} x^{3}+2 a x^{2}+a x+2$ , where a is a positive integer. It is given that 2 x+1 is a factor of p(x).

(a)

Find the value of a.

[ 3 ]

$a^{2}\left(-\frac{1}{2}\right)^{3}+2 a\left(-\frac{1}{2}\right)^{2}+a\left(-\frac{1}{2}\right)+2=0$ oe, soi

M1

Simplifies and solves for a

M1

FT their quadratic in a providing $\mathrm{p}\left(-\frac{1}{2}\right)=0$
oe attempted

a=4

A1

(b)

Hence factorise p(x).

[ 2 ]

$2(2 x+1)\left(4 x^{2}+1\right)$ mark final answer

M1 for $(2 x+1)\left(\frac{\text { their } a^{2}}{2} x^{2} \ldots+2\right)$
or $\left(8 x^{2}+2\right)$ found as quadratic factor

[Maximum number: 9]

$p(x)=ax^3+3x^2+bx-12$ has a factor of 2x+1. When p(x) is divided by x-3 the remainder is 105.

(a)

Find the value of a and of b.

[ 5 ]

$\begin{aligned}p\left(-\frac12\right)&:-\frac a8+\frac34-\frac b2-12=0\\p(3)&:27a+27+3b-12=105\end{aligned}$
M1

a+4b=-90
A1

9a+b=30
A1

$a=6,\quad b=-24$
M1, A1

(b)

Using your values of a and b, write p(x) as a product of 2 x+1 and a quadratic factor.

[ 2 ]

$(2x+1)(3x^2-12)$
B2; B1 for $3x^2$ , B1 for -12 and no extra x term.

(c)

Hence solve p(x)=0.

[ 2 ]

$x=-\frac12$
B1

$x=\pm2$
B1, dep on both B marks in part (b).

All question bank results loaded

Covered syllabus points

3. Factors of polynomials question 1

Question (a)

Question (b)

3. Factors of polynomials question 2

Question (a)

Question (b)

Question (c)