Question

Simplify the expression:

$\frac{(p^3+11p^2+28p)}{(p+4)}$

• A. $(p+7)$
• B. $p(p+7)$
• C. $(p+8)$
• D. $p(p+8)$
  

Answer 1

The correct option is B $p(p+7)$

$\text{Given:}\frac{(p^3+11p^2+28p)}{(p+4)}$
Taking 'p' common from the numerator, we get
$\frac{p(p^2+11p+28)}{(p+4)}..\left(i\right)$
Now, comparing the expression $p^2+11p+28$ with the identity $x^2+(a+b)x+ab$
$wenotethat,(a+b)=11andab=28.$
$So,$
$(7+4)=11and\left(7\right)\left(4\right)=28$

Hence,
$p^2+11p+28$
$=p^2+7p+4p+28$
$=p(p+7)+4(p+7)$
$=(p+4)(p+7)$
Therefore from (i),
$\frac{p(p^2+11p+28)}{(p+4)}=\frac{p(p+4)(p+7)}{(p+4)}=p(p+7)$