Question

Factor 5x 4 - 30x 2 - 135.

Answer 1

$5x^4 - 30x^2 - 135 \\ \\ GCF = 5 \\ \\ 5( \frac{5x^4}{5} + \frac{-30x^2}{5} - \frac{135}{5} ) \\ \\ 5(x^2 - 6x^2 - 27) \\ \\ 5(x^2 - 9)(x^2 + 3) \\ \\ 5(x^2 - 3^2)(x^2 + 3) \\ \\ 5(x + 3)(x - 3)(x^2 + 3)$

The final result is: 5(x + 3)(x - 3)(x² +3).

Answer 2

5x⁴ -30x² - 135

Factorizing 5x⁴ -30x² - 135,   Factoring out 5

5( x⁴ -6x² - 27)..............................(a)

Factoring ( x⁴ -6x² - 27). 

x⁴ -6x² - 27 = (x²)² -6(x²) - 27

Let y = x²

Therefore  (x²)² -6(x²) - 27 = y² -6y - 27

Therefore (a)  is the same as:  

5( x⁴ -6x² - 27) = 5(y² -6y - 27)...........(b)

y² -6y - 27 is a quadratic expression

Multiplying first and last terms = y² * -27 = -27y²

We think of two factors that multiply up to -27y², and add up to -6y which is the middle term of the quadratic expression.

The factors are :     -9y and +3y

Check:  (-9y)*(+3y) = -27y²,    +3y + (-9y) = 3y -9y = -6y 

In the expression y² -6y - 27, we replace the middle term -6y with (3y - 9y)

y² -6y - 27

y² +3y - 9y- 27

y(y+3) - 9(y+3)

(y+3)(y-9)

 y² -6y - 27 =  (y+3)(y-9) ........................(c)    

Recall y = x²

(y+3)(y-9) = (x²+3)(x²-9) ........................(d)

But (x²-9) is difference of two squares = x² - 3² = (x-3)(x+3)

(x²+3)(x²-9)

(x²+3)(x-3)(x+3) ...........................(e)

Therefore from (b)

5(x⁴ -6x² - 27) = 5(y² -6y - 27)

5(y² -6y - 27) = 5(y+3)(y-9)  =  5(x²+3)(x²-9) = 5(x²+3)(x-3)(x+3)

So 5x⁴ -30x² - 135  is =  5(x²+3)(x-3)(x+3)

I hope this helps.