Question

Show that 2x + 7 is a factor of 2x³ + 5x² – 11x – 14. Hence factorise the given expression completely, using the factor theorem. 

Answer 1

Let 2x + 7 = 0,
then 2x = -7
x = ${\displaystyle \frac{-7}{2}}$
substituting the value of x in f(x),
f(x) = 2x3 + 5x2 – 11x – 14

${\displaystyle f(-\frac{7}{2})=2{(-\frac{7}{2})}^3+5{(-\frac{7}{2})}^2-11(-\frac{7}{2})-14}$

= ${\displaystyle \frac{-343}{4}+\frac{245}{4}+\frac{77}{2}-14}$

= ${\displaystyle \frac{-343+245+154-56}{4}}$

= ${\displaystyle \frac{-399+399}{4}}$
= 0
Hence, (2x + 7) is a factor of f(x)
Proved.
Now, 2x³ + 5x² – 11x – 14
= (2x + 7)(x² – x – 2)
= (2x + 7)[x² – 2x + x – 2]
= (2x + 7)[x(x – 2) + 1(x – 2)]
= (2x + 7)(x + 1)(x – 2)

${\displaystyle 2x+7\text{)}\overline{2x^3+5x^2 –11x –14}(x^2– x –2}$
             2x³ +  7x²              
             –    –                       
                  – 2x²  – 11x
                  – 2x²  – 7x
                   +       +              
                            – 4x – 14
                            – 4x – 14
                            +     +       
                                x