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soldier1979 [14.2K]
3 years ago
8

(x^2−x−6x^2−8x+16)/(x2−x−12)

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
5 0

-3 and 4 cannot be the values of x in the given expression

The expression is given as:

\frac{(x^2-x-6x^2-8x+16)}{(x^2-x-12)}

Start by equating the denominator to 0

x^2 - x - 12 = 0

Expand the above equation

x^2 - 4x + 3x - 12 = 0

Factorize the equation

x(x - 4) + 3(x - 4) = 0

Factor out x - 4

(x + 3) (x - 4) = 0

Split

(x + 3)= 0\ or\  (x - 4) = 0

Remove brackets

x + 3= 0\ or\  x - 4 = 0

Solve for x

x =- 3\ or\  x = 4

Hence, -3 and 4 cannot be the value of x in the given expression

Read more about expressions at:

brainly.com/question/2972832

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