Standard +0.3 This question requires understanding that a common factor means both quadratics equal zero at x = -2, leading to two equations that can be manipulated algebraically. While it involves substitution and rearrangement, the steps are straightforward once the key insight is recognized, making it slightly easier than average.
Question 4:
4 | Explains how the factor theorem
applies with reference to
f(-2) = 0 for either function
or
Explains that either quadratic
expression can be factorised in
the form (x + 2) (x + p) as
(x + 2) is a factor
or
Explains that on division by
(x + 2) the remainder would be
zero | 2.4 | E1 | As ( x+2 ) is a factor, then when
x=−2, f(x) = 0
4−2b+c=0
4−2d +e=0
4−2b+c=4−2d +e
2d −2b=e−c
2 ( d −b )=e−c
Uses the factor theorem
with x=−2 substituted into one
of the expressions to obtain a
correct expression
NB It is not necessary to equate
to zero for this mark
or
Expands one of their factorised
forms and equates coefficients
correctly
(x+2)(x+ p)= x2 +(p+2)x+2p
p+2=b
2p =c
or
Divides one of the expressions by
(x + 2) to obtain a correct
remainder. Either one of
4−2b+c
4−2d +e | 1.1a | M1
Deduces both correct equations
using factor theorem or division
4−2b+c=0
4−2d +e=0
PI by 4−2b+c =4−2d +e
or
Expands both of their factorised
forms and equates coefficients to
deduce the correct equations –
must not use p in both | 2.2a | A1
Forms a single equation for
b, c, d and e and completes
rigorous argument to show the
required result
NB R1 can be awarded even if
E1 was not awarded | 2.1 | R1
Total | 4
Q | Marking instructions | AO | Mark | Typical solution
$x^2 + bx + c$ and $x^2 + dx + e$ have a common factor $(x + 2)$
Show that $2(d - b) = e - c$
Fully justify your answer.
[4 marks]
\hfill \mbox{\textit{AQA Paper 2 2019 Q4 [4]}}