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

2 Answers
Aug 10, 2016

=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)

Aug 10, 2016

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)"------------"