Question

How do you simplify `(2x^2-2)/(x^2-6x-7)*(x^2-10x+21)`?

Answer 1

`=2(x-1)(x-3)`

Explanation:

`(2x^2-2)/(x^2-6x-7)xx(x^2-10x+21)`

`=(2(x^2-1))/(x^2-7x+x-7)xx(x^2-7x-3x+21)`

`=(2(x+1)(x-1))/(x(x-7)+1(x-7))xx(x(x-7)-3(x-7))`

`=(2(x+1)(x-1))/((x-7)(x+1))xx(x-7)(x-3)`

`=2(x-1)(x-3)`

Answer 2

`2(x-1)(x-3)`

Explanation:

To simplify this expression, we require to `color(blue)"factorise"`

Let us begin with `2x^2-2` which has a common factor of 2 in both terms.
`color(blue)"-------------------------------------------------"`

`rArr2x^2-2=2(x^2-1)........ (A)`

Now `x^2-1` is a `color(red)"difference of squares"` which, in general factorises as follows.

`color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))`

here `x^2=(x)^2" and " 1=(1)^2rArra=x" and " b=1`

`rArrx^2-1=(x-1)(x+1)`

substitute back into (A)

`rArr2x^2-2=2(x-1)(x+1)`
`color(blue)"------------------------------------------------------"`

The standard form of a quadratic is `color(red)(|bar(ul(color(white)(a/a)color(black)(ax^2+bx+c)color(white)(a/a)|)))`

and to factorise consider the factors of the product ac, which must sum to give b.
`color(blue)"-----------------------------------------------------"`

For `x^2-6x-7 , ac=-7" and " b=-6`

The required values the are -7 and +1

`rArrx^2-6x-7=(x-7)(x+1)`

and for `x^2-10x+21 , ac=21" and " b=-10`

the required values are -7 and -3

`rArrx^2-10x+21=(x-7)(x-3)`
`color(blue)"----------------------------------------------------------"`

Substitute all of these factors back into the original expression.

`rArr(2(x-1)(x+1))/((x-7)(x+1))xx((x-7)(x-3))/1`

and 'cancelling' common factors between numerator/denominator

`=(2(x-1)cancel((x+1)))/(cancel((x-7))cancel((x+1)))xx(cancel((x-7))(x-3))/1`

=2(x-1)(x-3)
`color(blue)"------------"`