November 2021 Paper 1 Q16
16
\[\mathrm{f}(x) = 200x^3 + 100x^2 - 18x - 9\](a) Use the factor theorem to show that \((2x + 1)\) is a factor of \(\mathrm{f}(x)\). [2 marks]
(b) Hence solve \(\quad \mathrm{f}(x) = 0\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(200\left(-\dfrac{1}{2}\right)^3 + 100\left(-\dfrac{1}{2}\right)^2\) \(-\ 18\left(-\dfrac{1}{2}\right) - 9\) | M1 | oe eg \(200\left(-\dfrac{1}{8}\right) + 100\left(\dfrac{1}{4}\right) - 18\left(-\dfrac{1}{2}\right) - 9\) |
| \(-25 + 25 + 9 - 9 = 0\) with M1 seen | A1 | must evaluate each term and equate to zero |
Additional guidance
| Condone \(\left(\dfrac{1}{2}\right)^2\) for \(\left(-\dfrac{1}{2}\right)^2\) | |
| \(200\left(-\dfrac{1}{2}\right)^3 + 100\left(-\dfrac{1}{2}\right)^2 - 18\left(-\dfrac{1}{2}\right) - 9 = 0\) | M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| \((100x^2 - 9)\) | M1 | |
| \((10x - 3)(10x + 3)\) or \((x =)\ \sqrt{\dfrac{9}{100}}\) | M1dep | oe eg \((x =)\ \sqrt{0.09}\) |
| \(-0.5\) and \(-0.3\) and \(0.3\) | A1 | oe eg fractions |
Additional guidance
| \(-0.5\) and \(-0.3\) or \(-0.5\) and \(0.3\) with the other solution missing implies \((100x^2 - 9)\) | M1M0A0 |
| \(-0.3\) and \(0.3\) on answer line implies \((10x - 3)(10x + 3)\) | M2A0 |