18N.2.AHL.TZ0.H_6

pestleMathematicsAAHLPaper 218N· ahl-2-12-factor-and-remainder-theorems-sum-and-product-of-rootssource ↗

Let $P\left( x \right) = 2{x^4} - 15{x^3} + a{x^2} + bx + c$ , where $a$ , $b$ , $c \in \mathbb{R}$

Given that $\left( {x - 5} \right)$ is a factor of $P\left( x \right)$ , find a relationship between $a$ , $b$ and $c$ .

[2]

Given that ${\left( {x - 5} \right)^2}$ is a factor of $P\left( x \right)$ , write down the value of $P'\left( 5 \right)$ .

[1]

Given that ${\left( {x - 5} \right)^2}$ is a factor of $P\left( x \right)$ , and that $a = 2$ , find the values of $b$ and $c$ .

[3]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to substitute $x = 5$ and set equal to zero, or use of long / synthetic division (M1)

$2 \times {5^4} - 15 \times {5^3} + a \times {5^2} + 5b + c = 0$ A1

$\left( { \Rightarrow 25a + 5b + c = 625} \right)$

[2 marks]

0 A1

[1 mark]

EITHER

attempt to solve $P'\left( 5 \right) = 0$ (M1)

$\Rightarrow 8 \times {5^3} - 45 \times {5^2} + 4 \times 5 + b = 0$

$\left( {{x^2} - 10x + 25} \right)\left( {2{x^2} + \alpha x + \beta } \right) = 2{x^4} - 15{x^3} + 2{x^2} + bx + c$ (M1)

comparing coefficients gives $\alpha$ = 5, $\beta$ = 2

THEN

$b$ = 105 A1

$\therefore c = 625 - 25 \times 2 - 525$

$c$ = 50 A1

[3 marks]

Examiners’ report

[N/A]