Question

The polynomial p(x)=x^3-19x-30 has a known factor of (x+2)

Answer 1

Answer:

(x+2) (x+3) (x-5)

Step-by-step explanation:

x³-19x-30 = (x+2) (x²+ax-15)  ... x³=x*(1*x²)   while -30= (2)*(-15)

x³ + 0*x² - 19x -30 = x³ + (2+a)x² + (2a-15)x -30

2+a = 0

a = -2

x³-19x-30 = (x+2) (x²-2x-15) = (x+2) (x+3) (x-5)

Answer 2

Answer:

$(x + 2)\, (x + 3)\, (x - 5)$.

Step-by-step explanation:

Apply polynomial long division to divide $p(x)$ by the know factor, $(x + 2)$.

Fill in the omitted terms:

$\begin{aligned}p(x) &= x^{3}- 19\, x - 30 \\ &= x^{3} + 0\, x^{2} - 19\, x - 30\end{aligned}$.

The leading term of the dividend is currently $x^{3}$. On the other hand, the leading term of the divisor, $(x + 2)$, is $x$.

Hence, the next term of the quotient would be $x^{3} / x = x^{2}$.

$\begin{aligned}p(x) &= \cdots \\ &= x^{3} + 0\, x^{2} - 19\, x - 30 \\ &= x^{3} + 0\, x^{2} - 19\, x - 30 \\ &\quad - x^{2} \, (x + 2) + [x^{2} \, (x + 2)] \\ &= x^{3} + 0\, x^{2} - 19\, x - 30 \\ &\quad -x^{3} - 2\, x^{2} + [x^{2}\, (x + 2)] \\ &= -2\, x^{2} - 19\, x - 30 \\ &\quad + [x^{2} \, (x + 2)]\end{aligned}$.

The dividend is now $(-2\, x^{2} - 19\, x - 30)$, with $(-2\, x^{2})$ being the new leading term. The leading term of the divisor $(x + 2)$ is still $x$.

The next term of the quotient would be $(-2\, x^{2}) / x = -2\, x$.

$\begin{aligned}p(x) &= \cdots \\ &= -2\, x^{2} - 19\, x - 30 \\ &\quad + [x^{2} \, (x + 2)] \\ &=-2\, x^{2} - 19\, x - 30 \\ &\quad - (-2\,x ) \, (x + 2) + [(-2\, x) \, (x + 2)] + [x^{2} \, (x + 2)] \\ &= -2\, x^{2} - 19\, x - 30 \\ &\quad - (-2\, x^{2} - 4\, x) + [(x^{2} - 2\, x)\, (x + 2)] \\ &= -15\, x - 30 \\ &\quad + [(x^{2} - 2\, x)\, (x + 2)]\end{aligned}$.

The dividend is now $(-15\, x - 30)$, with $(-15\, x)$ as the new leading term.

The next term of the quotient would be $(-15\, x) / x = -15$.

$\begin{aligned}p(x) &= \cdots \\ &= -15\, x - 30 \\ &\quad + [(x^{2} - 2\, x)\, (x + 2)] \\ &=-15\, x - 30 \\ &\quad -(-15)\, (x + 2) + [(-15)\, (x + 2)] + [(x^{2} - 2\, x)\, (x + 2)] \\ &= -15\, x - 30 \\ &\quad -(-15\, x - 30) + [(x^{2} - 2\, x - 15)\, (x + 2)] \\ &= (x^{2} - 2\, x - 15)\, (x + 2)\end{aligned}$.

In other words:

$\displaystyle \text{ \frac{x^{3} - 19\, x - 30}{x + 2} = x^{2} - 2\, x - 15 given that x+2 \ne 0 }$.

The next step is to factorize the quadratic polynomial $(x^{2} - 2\, x - 15)$.

Apply the quadratic formula to find the two roots of $x^{2} - 2\, x - 15 = 0$:

$\begin{aligned} x_{1} &= \frac{-(-2) + \sqrt{2^{2} - 4\times 1 \times (-15)}}{2} \\ &= \frac{2 + 8}{2}= 5\end{aligned}$.

$\begin{aligned} x_{2} &= \frac{2 - 8}{2} = -3\end{aligned}$.

By the factor theorem, the two factors of $(x^{2} - 2\, x - 15)$ would be $(x - 5)$ and $(x - (-3))$. That is:

$x^{2} - 2\, x -15 = (x + 3)\, (x - 5)$.

Therefore:

$\begin{aligned}p(x) &= \cdots \\ &= (x^{2} - 2\, x - 15)\, (x + 2) \\ &= (x + 2)\, (x + 3)\, (x - 5)\end{aligned}$.