Using the factor theorem, show that $x+3$ is a factor of $y(x)$ in the polynomial $y(x)=x^3 + 5x^2 + 8x + 6$

Question

Considering the cubic polynomial

$$y(x)=x^3 + 5x^2 + 8x + 6,$$

show that $x+3$ is a factor of $y(x)$ using the factor theorem.

Sorry, I'm a bit stuck with this question. Would you use long division? Or would that be for the second part of the question where I have to fully factorise $y(x)$ and hence show that $x=−3$ is the only real zero of $y(x)$.?

Answer 1

Since $x+3$ is a factor of $y(x)$

$y(-3)=0$ $$(-3)^3+5(-3)^2+8(-3)+6=0$$ $$-27+45-24+6=0$$ $$0=0$$ Hence $x+3$ is a factor of $y(x)$

Answer 2

If $x+3$ is a factor, this means that: $$x^3+5x^2+8x+6=(x+3)(\text{some quadratic})$$ From this you can take from what you already know that when $x=-3$, $y(x)=0$

So plug $-3$ into your equation for: $$(-3)^3+5(-3)^2+8(-3)+6$$ $$\to -27+45-24+6$$ $$\to -51+51$$ $$\to 0$$ and therefore it is a factor.

With regards to the second part, you have to find the thing I've labelled "$(\text{some quadratic})$"

Use that: $$(x+3)(Ax^2+Bx+C)=x^3+5x^2+8x+6$$expand and compare to see that: $$Ax^3=x^3\to A=1$$ $$3Ax^2+Bx^2=5x^2\to B=2$$ $$3Bx+Cx=8x\to C=2$$ $$3C=6\to C=2\space \text{(confirmation)}$$

Thus: $$(x+3)(x^2+2x+2)=x^3+5x^2+8x+6$$In order to show that $x=-3$ is the only real root, prove that the roots of $x^2+2x+2$ are not real. We use the discriminant $D=B^2-4AC$ for this. If there are no real roots, then $D<0$ (the reason why this is true is due to the quadratic formula: $x=\frac{-B\pm\sqrt{D}}{2A}$, we cannot square root a negative value and expect a real answer . Hence check this: $$B^2-4AC\to 2^2-4(1)(2)\to 4-8\to -4<0$$ which therefore means it has no real roots.