Factoring (six polynomials)

Text

Factor: (a) $81a^6-64b^4$; (b) $12c^2d^4-75c^2d^2$; (c) $x^4+6x^2+9$; (d) $a^2+3a+2$; (e) $a^2-13a-14$; (f) $24a^2-54b^2$.

Solution

(a) A difference of squares: $81a^6=(9a^3)^2$ and $64b^4=(8b^2)^2$. $81a^6-64b^4=(9a^3)^2-(8b^2)^2=(9a^3-8b^2)(9a^3+8b^2).$

(b) Factor out the common term $3c^2d^2$: $12c^2d^4-75c^2d^2=3c^2d^2(4d^2-25).$ Now $4d^2-25=(2d)^2-5^2$ is a difference of squares: $3c^2d^2(4d^2-25)=3c^2d^2(2d-5)(2d+5).$

(c) A perfect square of a binomial: $x^4=(x^2)^2$, $9=3^2$ and $6x^2=2\cdot x^2\cdot 3$. $x^4+6x^2+9=(x^2)^2+2\cdot x^2\cdot 3+3^2=(x^2+3)^2.$

(d) Characteristic trinomial: two numbers with sum $3$ and product $2$, namely $1$ and $2$. $a^2+3a+2=(a+1)(a+2).$

(e) Characteristic trinomial: two numbers with sum $-13$ and product $-14$, namely $-14$ and $1$. $a^2-13a-14=(a-14)(a+1).$

(f) Factor out $6$: $24a^2-54b^2=6(4a^2-9b^2).$ Now $4a^2-9b^2=(2a)^2-(3b)^2$ is a difference of squares: $6(4a^2-9b^2)=6(2a-3b)(2a+3b).$

$\boxed{(9a^3-8b^2)(9a^3+8b^2)\ ;\ \ 3c^2d^2(2d-5)(2d+5)\ ;\ \ (x^2+3)^2\ ;\ \ (a+1)(a+2)\ ;\ \ (a-14)(a+1)\ ;\ \ 6(2a-3b)(2a+3b)}$