Moderate -0.8 This is a straightforward application of the Remainder Theorem requiring substitution and basic algebraic manipulation. Part (a) is immediate substitution, part (b) uses f(-2) = -5 to find k, and part (c) is routine factorization once k is known. All steps are standard textbook exercises with no problem-solving insight required.
2.
$$f ( x ) = ( 2 x - 3 ) ( x - k ) - 12$$
where \(k\) is a constant.
a.Write down the value of \(\mathrm { f } ( k )\)
When \(\mathrm { f } ( x )\) is divided by ( \(x + 2\) ) the remainder is - 5
b. find the value of \(k\).
c. Factorise \(\mathrm { f } ( x )\) completely.
**(a)** | $f(k) = -12$ | B1 | For substituting $x = k$ and $f(k) = (2k-3)(k-k) - 12 = -12$ |
**(b)** | $k = -1$ | A1 | M1 for substituting $x = -2$ and equating to $-5$ to form an equation in $k$ and solving to find $k$. E.g. $f(-2) = (2(-2)-3)(-2-k)-12 = -5 \Rightarrow (-7)(-2-k)-12 = -5 \Rightarrow -7(-2-k) = 7 \Rightarrow -2-k = -1 \Rightarrow k = \ldots$ |
**(c)** | $(2x+5)(x-3)$ | A1 | M1 for multiplying and substituting their constant value of $k$ (in either order). The multiplying-out may occur earlier. E.g. $(2x-3)(x+1)-12 = 2x^2 - x - 3 - 12 = 2x^2 - x - 15$. M1 for an attempt to factorise their three term quadratic e.g. $2x^2 - x - 15 = (2x \pm 5)(x \pm 3)$. The correct answer, as a product of factors, is required. |
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2.
$$f ( x ) = ( 2 x - 3 ) ( x - k ) - 12$$
where $k$ is a constant.\\
a.Write down the value of $\mathrm { f } ( k )$
When $\mathrm { f } ( x )$ is divided by ( $x + 2$ ) the remainder is - 5\\
b. find the value of $k$.\\
c. Factorise $\mathrm { f } ( x )$ completely.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q2 [6]}}