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:
[tex](x + 2)\, (x + 3)\, (x - 5)[/tex].
Step-by-step explanation:
Apply polynomial long division to divide [tex]p(x)[/tex] by the know factor, [tex](x + 2)[/tex].
Fill in the omitted terms:
[tex]\begin{aligned}p(x) &= x^{3}- 19\, x - 30 \\ &= x^{3} + 0\, x^{2} - 19\, x - 30\end{aligned}[/tex].
The leading term of the dividend is currently [tex]x^{3}[/tex]. On the other hand, the leading term of the divisor, [tex](x + 2)[/tex], is [tex]x[/tex].
Hence, the next term of the quotient would be [tex]x^{3} / x = x^{2}[/tex].
[tex]\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}[/tex].
The dividend is now [tex](-2\, x^{2} - 19\, x - 30)[/tex], with [tex](-2\, x^{2})[/tex] being the new leading term. The leading term of the divisor [tex](x + 2)[/tex] is still [tex]x[/tex].
The next term of the quotient would be [tex](-2\, x^{2}) / x = -2\, x[/tex].
[tex]\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}[/tex].
The dividend is now [tex](-15\, x - 30)[/tex], with [tex](-15\, x)[/tex] as the new leading term.
The next term of the quotient would be [tex](-15\, x) / x = -15[/tex].
[tex]\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}[/tex].
In other words:
[tex]\displaystyle \text{$\frac{x^{3} - 19\, x - 30}{x + 2} = x^{2} - 2\, x - 15$ given that $x+2 \ne 0$}[/tex].
The next step is to factorize the quadratic polynomial [tex](x^{2} - 2\, x - 15)[/tex].
Apply the quadratic formula to find the two roots of [tex]x^{2} - 2\, x - 15 = 0[/tex]:
[tex]\begin{aligned} x_{1} &= \frac{-(-2) + \sqrt{2^{2} - 4\times 1 \times (-15)}}{2} \\ &= \frac{2 + 8}{2}= 5\end{aligned}[/tex].
[tex]\begin{aligned} x_{2} &= \frac{2 - 8}{2} = -3\end{aligned}[/tex].
By the factor theorem, the two factors of [tex](x^{2} - 2\, x - 15)[/tex] would be [tex](x - 5)[/tex] and [tex](x - (-3))[/tex]. That is:
[tex]x^{2} - 2\, x -15 = (x + 3)\, (x - 5)[/tex].
Therefore:
[tex]\begin{aligned}p(x) &= \cdots \\ &= (x^{2} - 2\, x - 15)\, (x + 2) \\ &= (x + 2)\, (x + 3)\, (x - 5)\end{aligned}[/tex].
