How to factor $2x^3 + 21x^2 +27x$

Question

I am having trouble with factoring $2x^3 + 21x^2 +27x$. The answer is $x(x+9)(2x+3)$ but not sure how that was done. Obviously I factored out the $x$ to get $x(2x^2+21x+27)$ then from there I am lost. I tried the AC method and grouping. Can someone show the steps? Thanks!

Answer 1

Quite simple; The AC Method can be used here:

$$2x^3 + 21x^2 +27x =$$ $$x(2x^2 + 21x +27) =$$ Now we have to find two factors that multiply to $54$ ($2 \times 27$), and add up to $21$. Two such numbers are $3$ and $18$, which are now used to split apart the polynomial. $$x((2x^2 + 3x) + (18x + 27)) =$$ From here it is simple grouping. $$x(x(2x + 3) + 9(2x + 3)) =$$ $$x(x + 9)(2x + 3)$$

Answer 2

$$2x^3+21x^2+27x\\=x(2x^2+21x+27)\\=\dfrac{x}{2}(4x^2+42x+54)\\=\dfrac{x}{2}(2x+18)(2x+3)\\=x(x+9)(2x+3)$$for factorizing $4x^2+42x+54$ we know that it must in form of $(2x+a)(2x+b)$. So:$$4x^2+42x+54=(2x+a)(2x+b)=4x^2+(a+b)2x+ab$$which implies$$ab=54\\a+b=21$$clearly the only numbers satisfying those equalities are $a=18$ and $b=3$ or vice versa.

Another way to attain such numbers when guessing them is not that easy is to solve the following equation$$4x^2+42x+54=0$$or$$2x^2+21x+27=0$$which leads to $$x=\dfrac{-21\pm\sqrt{{21^2-4\times 2\times 27}}}{4}=\dfrac{-21\pm 15}{4}$$therefore$$x_1=\dfrac{-21-15}{4}=-9\\x_2=-\dfrac{3}{2}$$so$$2x^2+21x+27=2(x+9)(x+\dfrac{3}{2})=(x+9)(2x+3)$$

Answer 3

Find the roots $x_1$ and $x_2$ of $2x^2+21x+27$ with the standard abc method.

Then $2x^2+21x+27=c(x-x_1)(x-x_2)$ with an easily computable $c$

Answer 4

1)

$2x^3 + 21x^2 +27x = x(2x^2 + 21x + 27) = x(2x(x + 9) + 3x + 27) =x(2x(x+9) + 3(x + 9)) = x(2x + 3)(x+9)$.

2)

$2x^3 + 21 x^2 + 27x = x(2x^2 + 21x + 27) = x*2*(x - a)(x+b)$ where $a,b$ are solutions to $2x^2 + 21x + 27=0$. i.e. $x = \frac {-21 \pm {21^2 - 8*27}}{4} = \frac {-21 \pm \sqrt {225}}4 = \frac {-21 \pm 15}4 = -9, -\frac 32$ so

$2x^3 + 21 x^2 + 27x = x*2*(x + \frac 32)(x + 9) = x(2x +3)(x+9)$.

3)

$2x^3 + 21x^2 + 27x = (ax + b)(cx + d)(ex + f)$ and

$ace = 2$. Wolog, $a = 2$ and $c, e = 1$.

So $2x^3 + 21x^2 + 27x = (2x + b)(x + d)(x + f)$

$2f + 2d + b = 21$

$2df + bf + bd = 27$

$bdf = 0$

$b$ is odd so $b\ne 0$. Wolog $d= 0$.

$2f + b = 21$

$bf = 27$

So $b=3; f= 9$

So $x(2x + 3)(x + 9)$.