Question

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

Rewrite p(x) as a product of linear factors

Answer 1

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

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

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

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{$(x^(3) - 19\, x - 30)/(x + 2) = x^(2) - 2\, x - 15$ given that $x+2 \\e 0$}`.

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* 1 * (-15)}}{2} \\ &= (2 + 8)/(2)= 5\end{aligned}`.

`\begin{aligned} x_(2) &= (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}`.