How do you simplify #\frac { 6x ^ { 2} - 42x + 72} { 4x ^ { 2} - 24x + 32}#?

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Answer 1 · 2017-03-26T08:44:42

#(3(x-3))/(2(x-2))#

First, you would factor both the polynomials:

#(cancel6^3(x^2-7x+12))/(cancel4^2(x^2-6x+8))#

and then find the zeros of the polynomials by the quadratic formula:

N. #->x^2-7x+12#

#x=(7+-sqrt(49-48))/2=(7+-1)/2#

the zeros are 3 and 4, and you factor:

#3(x^2-7x+12)=3(x-3)(x-4)#

D.#->x^2-6x+8#

#x=3+-sqrt(9-8)=3+-1# (applying the simple quadratic formula)

the zeros are 2 and 4, and you factor:

#2(x^2-6x+8)=2(x-2)(x-4)#

Finally you simplify:

#(3(x^2-7x+12))/(2(x^2-6x+8))=(3(x-3)cancel((x-4)))/(2(x-2)cancel((x-4)))# and get:

#(3(x-3))/(2(x-2))#