-3 and 4 cannot be the values of x in the given expression
The expression is given as:
[tex]\frac{(x^2-x-6x^2-8x+16)}{(x^2-x-12)}[/tex]
Start by equating the denominator to 0
[tex]x^2 - x - 12 = 0[/tex]
Expand the above equation
[tex]x^2 - 4x + 3x - 12 = 0[/tex]
Factorize the equation
[tex]x(x - 4) + 3(x - 4) = 0[/tex]
Factor out x - 4
[tex](x + 3) (x - 4) = 0[/tex]
Split
[tex](x + 3)= 0\ or\ (x - 4) = 0[/tex]
Remove brackets
[tex]x + 3= 0\ or\ x - 4 = 0[/tex]
Solve for x
[tex]x =- 3\ or\ x = 4[/tex]
Hence, -3 and 4 cannot be the value of x in the given expression
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