Question

Factorise: (x^2-4x) (x^2-4x-1) -20​

Answer 1

Answer: $(x-5)(x+1)(x-2)^{2}$

Step-by-step explanation:

$(x^{2} - 4x)(x^{2} -4x -1) - 20$

Let  $x^{2} - 4x$  be  = a

$(a)(a -1) - 20$

$=a^{2} -a - 20$

Now, factorise it using middle term splitting.

$= a^{2} -5a + 4a - 20$

$= a(a - 5) + 4(a - 5)$

$= (a-5)(a+4)$

Now, put the value of 'a'.

$= (x^{2} - 4x -5)(x^{2} - 4x + 4)$

Now, factorise the two quadratic expressions in the brackets by middle term splitting. You will get 4 factors (2 factors of each expression).

$= [x^{2} - 5x + x -5][x^{2} - 2x -2x + 4]$

$= [x(x-5) + 1(x -5)][x(x-2)-2(x -2)]$

$= (x-5)(x+1)(x-2)(x-2)$

$= (x-5)(x+1)(x-2)^{2}$

Answer 2

Answer:

$x = - 1$

Step-by-step explanation:

Given equation is

$( {x}^{2} - 4x)( {x}^{2} - 4x - 1) - 20$

${x}^{4} - 4 {x}^{3} - {x}^{2} - 4 {x}^{3} + 16 {x}^{2} + 4x - 20$

${x}^{4} - 8 {x}^{3} + 15 {x}^{2} + 4x - 20$

The value of x can be (Factors of constant/Factors of coefficient of x^4)

Factors of 1 : 1

Factors of 20: 1, 2, 4, 5, 10, 20

So the value of x may be in this list

(1/1), (2/1), (4/1), (5/1), (10/1), (20/1)

And the values of x may be the negatives of above numbers

Lets check out this one

put x= 1

1-8+15+4-20 = 20-8-20 = 20-28 = -8

This is not equal to zero

put x=-1

1+8+15-4-20 = 24-4-20 = 24-24 = 0

The solution of x is -1 , 5 , 2 , 2