0
$\begingroup$

$$125x^3 + 216$$

I have tried to factor it but because the square root of $216$ is a decimal, I can't figure out how to do the problem.

$\endgroup$
2
  • 1
    $\begingroup$ Note that you have a 3rd power; the squareroot is irrelevant. $\endgroup$
    – quid
    Dec 28, 2014 at 22:48
  • 4
    $\begingroup$ Notice that $216=6^3$. $\endgroup$
    – Bill Cook
    Dec 28, 2014 at 22:48

4 Answers 4

9
$\begingroup$

Hint: $$A^3+B^3=(A+B)(A^2-AB+B^2)$$

where $125x^3=A^3$ and $B^3=216$, then $A=?$ and $B=?$...

$\endgroup$
3
$\begingroup$

We start by observing that $216 = 6^3$ and $125=5^3$, so we might consider $x=-6/5$.

We have $$125(-6/5)^3 + 216 = 125 \cdot(-216/125) + 216 = 0$$ so by the factor theorem, one root is $5x+6$.

Given this first factor, we do the polynomial division and obtain

$$125x^3 + 216 = (5x+6)(25x^2-30x+36)$$

Obviously we can't reduce the linear factor any further, but what about the quadratic factor?

Again, we want to use the factor theorem: if we can find a root, then we have another linear factor. So are there any roots of this quadratic?

No. The discriminant is $$\Delta = 30^2 - 4 \cdot 25 \cdot 36 = -2700<0$$ and hence there are no real roots.

It follows that the factorisation is

$$125x^3 + 216 = (5x+6)(25x^2-30x+36)$$

$\endgroup$
2
$\begingroup$

(5x+6)(25x^2-30x+36) You should check for perfect cubes, not squares.

$\endgroup$
1
$\begingroup$

Note that $125x^3+216=(5x)^3+6^3, $ then we have $(5x+6)((5x)^2-5x\cdot6+6^2)$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .