Question

Simplify the following -

i) $3n^2-2n\left(n.3\right)+3n(2n+3)$

ii) $pq(p-q)+p^2(p^2-q^2)$

iii) $\frac{2}{7}(14n^2-21y^2)(-5n^2-3y^2)$

iv) $(n^2+5n+3)(n+5)$

Answer 1

$\left(A\right)3n^2-2n\left(n.3\right)+3n(2n+3)$

$=3n^2-6n^2+6n^2+9n$

$=3n^2+9n$

$\left(B\right)pq(p-q)+p^2(p^2-q^2)$

$=pq(p-q)+p^2(p+q)(p-q)$

$=(p-q)(pq+p^3+p^{2}q)$

or

$=p(p-q)(q+p^2+pq)$

$\left(C\right)\frac{2}{7}(14n^2-21y^2)(-5n^2-3y^2)$

$=2(\frac{14}{2}{n}^2-\frac{21}{7}{y}^2)(-5n^2+3y^2)$

$=-2(2n^2-3y^2)(5n^2+3y^2)$

$-2(10n^4+6n^2{y}^2-15n^2{y}^2-9y^4)$

$=-20n^4+18n^2{y}^2+18y^4$

$\left(D\right)(n^2+5n+3)(n+5)$

$=n^3+5n^2+5n^2+25n+3n+15$

$=n^3+{10}^2+28n+15$