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

2 Answers
Aug 10, 2016

Answer:

#=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

Answer:

#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)"------------"#