Question

(x^2−x−6x^2−8x+16)/(x2−x−12)

Answer 1

-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

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