Question

Simplify completely: the quantity 8 times x to the third power plus 6 times x to the 2nd power minus 4 times x all over negative 2 times x
a.−4x^2 − 3x + 2

b.−4x^2 + 3x + 2

c.4x^4 + 3x^3 − 2x^2

d.4x^2 + 6x^2 − 4x

Answer 1

The answer is a.−4x^2 − 3x + 2

the quantity 8 times x to the third power plus 6 times x to the 2nd power minus 4 times x all over negative 2 times x is: $\frac{8 x^{3}+6 x^{2} -4x }{-2x}$

$\frac{8 x^{3}+6 x^{2} -4x }{-2x} = -\frac{8 x^{3}+6 x^{2} -4x }{2x}=- \frac{2x*4 x^{2} +2x*3x-2x*2}{2x} =- \frac{2x(4x^{2} +3x-2)}{2x} = \\ \\ =-(4x^{2} +3x-2)=-4x^{2} -3x+2$