How do you simplify (2x^2-2)/(x^2-6x-7)*(x^2-10x+21)?
2 Answers
Explanation:
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 acolor(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)"------------"