Question

x⁴-2x³-7x²+8x+12/x-3​

Answer 1

${x}^{4} - {2x}^{3} - {7x}^{2} + 8x + 12 \\ = {x}^{4} - {3x}^{3} + {x}^{3} - {3x}^{2} - {4x}^{2} + 12x - 4x + 12 \\ = {x}^{3} (x - 3) + {x}^{2} (x - 3) - 4x(x - 3) - 4(x - 3) \\ = (x - 3)( {x}^{3} + {x}^{2} - 4x - 4)$

So,, the Quotient is,,x^3+x^2-4x-4

and,, the remainder is,,zero

Hope it helps you.. please mark it as Brainliest answer.... thanks...

Answer 2

Answer:

x^4-2x^3-7x^2+8x+12

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

=x^3 (x-3)+x^2 (x-3)-4x (x-3)-4 (x-3)

=(x-3)(x^3+x^2-4x-4)

=(x-3)(x^3+2x^2-x^2-2x-2x-4)

=(x-3)[x^2 (x+2)-x (x+2)-2 (x+2)]

=(x-3)(x+2)(x^2-x-2)

=(x-3)(x+2)(x^2+x-2x-2)

=(x-3)(x+2)[x (x+1)-2 (x+1)]

=(x-3)(x+2)(x+1)(x-2)

This is your required answer

Step-by-step explanation: