| Date | November 2017 | Marks available | 3 | Reference code | 17N.1.hl.TZ0.3 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Hence or otherwise and Factorize | Question number | 3 | Adapted from | N/A |
Question
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]
a.
Hence or otherwise, factorize \(q(x)\) as a product of linear factors.
[3]
b.
Markscheme
\(q(4) = 0\) (M1)
\(192 - 176 + 4k + 8 = 0{\text{ }}(24 + 4k = 0)\) A1
\(k = - 6\) A1
[3 marks]
a.
\(3{x^3} - 11{x^2} - 6x + 8 = (x - 4)(3{x^2} + px - 2)\)
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]
b.
Examiners report
[N/A]
a.
[N/A]
b.
