Question

factorise=9p^4-24p^2q^2+16q^2-256r^2

Answer 1

Answer:

The factors of the given polynoial are

$\bf\,(3p^2-4q^2+16r^2)(3p^2-4q^2-16r^2)$

Step-by-step explanation:

The given polynomial is

$9p^4-24p^2q^2+16q^2-256r^2$

The given polynomial is factorized by using suitabable algebraic identities

$9p^4-24p^2q^2+16q^2-256r^2$

$=[(3p^2)^2-2\times\,3p^2\,4q^2+(4q^2)^2]-\,(16r^2)^2$

Using, the identity

$\boxed{\bf\,a^2-2ab+b^2=(a-b)^2}$

$=(3p^2-4q^2)^2-(16r^2)^2$

Using, the identity

$\boxed{\bf\,a^2-b^2=(a+b)(a-b)}$

$=(3p^2-4q^2)^2-(16r^2)^2$

$=(3p^2-4q^2+16r^2)(3p^2-4q^2-16r^2)$

Answer 2

(3$p^{2}$- 4q + 16r)  (3$p^{2}$- 4q - 16r)

Step-by-step explanation:

Given,

9$p^{4}$ - 24$p^{2}$$q^{2}$ + 16$q^{2}$ - 256$r^{2}$

Grouping into two separate equations:

(9$p^{4}$ - 24$p^{2}$$q^{2}$ + 16$q^{2}$) - (256$r^{2}$)

Finding the perfect squares:

(($3p^{2}$$)^{2}$ - 24$p^{2}$$q^{2}$ + (4$q$$)^{2}$) - (256$r^{2}$)

Splitting the middle as 2ab

(($3p^{2}$$)^{2}$ - (2×3$p^{2}$×4q) + (4$q$$)^{2}$) - (256$r^{2}$)

Applying the formula of: $a^{2}$ - 2ab - $b^{2}$ = $(a-b)^{2}$

(3$p^{2}$- 4q$)^{2}$  - (256$r^{2}$)

Finding the perfect square of 256$r^{2}$:

(3$p^{2}$- 4q$)^{2}$ - (16r$)^{2}$

Applying the formula of: $a^{2}$ - $b^{2}$ = (a+b) (a-b)

(3$p^{2}$- 4q + 16r)  (3$p^{2}$- 4q - 16r)

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