4. $x^{2} + bx + c$ and $x^{2} + dx + e$ have a common factor $(x + 2)$ - AQA - A-Level Maths Pure - Question 4 - 2019 - Paper 2
Question 4
4.
$x^{2} + bx + c$ and $x^{2} + dx + e$ have a common factor $(x + 2)$.
Show that $2(d - b) = e - c$.
Fully justify your answer.
Worked Solution & Example Answer:4. $x^{2} + bx + c$ and $x^{2} + dx + e$ have a common factor $(x + 2)$ - AQA - A-Level Maths Pure - Question 4 - 2019 - Paper 2
Step 1
Show that $(x + 2)$ is a factor.
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Answer
By applying the factor theorem, if (x+2) is a factor of the quadratic expressions, then substituting x=−2 into either expression must yield zero.
Substituting into the first expression, we get:
(−2)2+b(−2)+c=0
which simplifies to:
4−2b+c=0.
This can be rearranged to give:
c=2b−4.
Step 2
Substituting $x = -2$ into the second quadratic.
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Answer
Now, substituting x=−2 into the second quadratic, we have:
(−2)2+d(−2)+e=0
which simplifies to:
4−2d+e=0.
Rearranging gives:
e=2d−4.
Step 3
Finding the relation between $e$, $c$, $d$, and $b$.
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Answer
We now have two expressions:
c=2b−4
e=2d−4
To find the relationship between e, c, d, and b, we can equate both expressions to eliminate 4:
Substituting c in the equation:
e−c=(2d−4)−(2b−4)
This simplifies to:
e−c=2d−2b
Rearranging results in:
2(d−b)=e−c.
