17N.1.AHL.TZ0.H_3

pestleMathematicsAAHLPaper 117N· ahl-2-12-factor-and-remainder-theorems-sum-and-product-of-rootssource ↗

Consider the polynomial $q(x) = 3{x^3} - 11{x^2} + kx + 8$ .

Given that $q(x)$ has a factor $(x - 4)$ , find the value of $k$ .

[3]

Hence or otherwise, factorize $q(x)$ as a product of linear factors.

[3]

Markscheme / solution

$q(4) = 0$ (M1)

$192 - 176 + 4k + 8 = 0{\text{ }}(24 + 4k = 0)$ A1

$k = - 6$ A1

[3 marks]

$3{x^3} - 11{x^2} - 6x + 8 = (x - 4)(3{x^2} + px - 2)$

equate coefficients of ${x^2}$ : (M1)

$- 12 + p = - 11$

$p = 1$

$(x - 4)(3{x^2} + x - 2)$ (A1)

$(x - 4)(3x - 2)(x + 1)$ A1

Note: Allow part (b) marks if any of this work is seen in part (a).

Note: Allow equivalent methods (eg, synthetic division) for the M marks in each part.

[3 marks]

Examiners’ report

[N/A]