Why does $3x^4 + 16x^3 + 20x^2 - 9x - 18$ = $(x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3$? $$3x^4 + 16x^3 + 20x^2 - 9x - 18 $$
When simplified I arrive to:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6}) $$
But the math book wrote:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3 $$
with that extra 3 at the end. The graph calculator seem to agree with that extra 3 as well. what did I do wrong?
 A: Let $ a\ne 0$.
If the equation $$ax^2+bx+c=0$$ has two roots $ x_1 $ and $ x_2 $, then
$$ax^2+bx+c=a(x-x_1)(x-x_2)$$
do not forget the coefficient $ a$.
A: By the rational root test the polynomial has the two roots $x=-2$ and $x=-3$ so that
$$
3x^4 + 16x^3 + 20x^2 - 9x - 18=(x+2)(x+3)(3x^2+x-3)
$$
The extra $3$ is not an exponent but is the leading coefficient in $3x^2+x-3$.
A: For any real $k$ we have
$$3x^4+16x^3+20x^2-9x-18=3\left(x^4+\frac{16}{3}x^3+\frac{20}{3}x^2-3x-6\right)=$$
$$=3\left(\left(x^2+\frac{8}{3}x+k\right)^2-\frac{64}{9}x^2-\frac{16k}{3}x-2kx^2-k^2+\frac{20}{3}x^2-3x-6\right)=$$
$$=3\left(\left(x^2+\frac{8}{3}x+k\right)^2-\left(\left(2k+\frac{4}{9}\right)x^2+\left(\frac{16k}{3}+3\right)x+k^2+6\right)\right).$$
Now, we'll choose $k$ such that $$2k+\frac{4}{9}>0$$ and
$$\left(\frac{16k}{3}+3\right)^2-4\left(2k+\frac{4}{9}\right)(k^2+6)=0,$$
which gives $k=\frac{5}{2}.$
Id est, $$3x^4+16x^3+20x^2-9x-18=3\left(\left(x^2+\frac{8}{3}x+\frac{5}{2}\right)^2-\left(\frac{49}{9}x^2+\frac{49}{2}x+\frac{49}{4}\right)\right)=$$
$$=3\left(\left(x^2+\frac{8}{3}x+\frac{5}{2}\right)^2-\left(\frac{7}{3}x+\frac{7}{2}\right)^2\right)=3(x^2+5x+6)\left(x^2+\frac{x}{3}-1\right)=$$
$$=3(x+2)(x+3)\left(x-\frac{-\frac{1}{3}+\sqrt{\frac{37}{9}}}{2}\right)\left(x-\frac{-\frac{1}{3}-\sqrt{\frac{37}{9}}}{2}\right)=$$
$$=3(x+2)(x+3)\left(x-\frac{-1+\sqrt{37}}{6}\right)\left(x-\frac{-1-\sqrt{37}}{6}\right).$$
