Question
Show that $2x + 7$ is a factor of $2x^3 + 5x^2 – 11x – 14$. Hence factorise the given expression completely, using the factor theorem.
✓
Answer
Let $2x + 7 = 0,$
then $2x = -7$
$x=\frac{-7}{2}$
substituting the value of x in f(x),
$f(x) = 2x^3 + 5x^2 – 11x – 14$
$f\left(-\frac{7}{2}\right)=2\left(-\frac{7}{2}\right)^3+5\left(-\frac{7}{2}\right)^2-11\left(-\frac{7}{2}\right)-14$
$=\frac{-343}{4}+\frac{245}{4}+\frac{77}{2}-14$
$=\frac{-343+245+154-56}{4}$
$=\frac{-399+399}{4}$
$=0$
Hence, (2x + 7) is a factor of f(x)
Proved.
Now, $2x^3 + 5x^2 – 11x – 14$
$=(2 x+7)\left(x^2-x-2\right)$
$=(2 x+7)\left[x^2-2 x+x-2\right]$
$=(2 x+7)[x(x-2)+1(x-2)]$
$=(2 x+7)(x+1)(x-2)$
then $2x = -7$
$x=\frac{-7}{2}$
substituting the value of x in f(x),
$f(x) = 2x^3 + 5x^2 – 11x – 14$
$f\left(-\frac{7}{2}\right)=2\left(-\frac{7}{2}\right)^3+5\left(-\frac{7}{2}\right)^2-11\left(-\frac{7}{2}\right)-14$
$=\frac{-343}{4}+\frac{245}{4}+\frac{77}{2}-14$
$=\frac{-343+245+154-56}{4}$
$=\frac{-399+399}{4}$
$=0$
Hence, (2x + 7) is a factor of f(x)
Proved.
Now, $2x^3 + 5x^2 – 11x – 14$
$=(2 x+7)\left(x^2-x-2\right)$
$=(2 x+7)\left[x^2-2 x+x-2\right]$
$=(2 x+7)[x(x-2)+1(x-2)]$
$=(2 x+7)(x+1)(x-2)$
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