Question

x+2 x²-x-6
___÷ _____
4y 12y²

Answer 1

Step-by-step explanation:

x={-2, 3}

PREMISES

x^2-x-6=0

CALCULATIONS

x^2-x-6=0

(x+2)(x-3)=0

Either x+2=0, x=-2 and/or x-3=0, x=3

x=-2, 3

PROOF

If x={-2, 3}, then the equations

x^2-x-6=0

[(-2)^2-(-2)-6]-[3^2–3–6]=0

(4+2–6)-(9–9)=0

0–0=0 and

0=0 prove the roots (zeros) x={-2, 3} of the polynomial x^2-x-6=0