# Cubic Equation. (Factorisation)

I'm given this question, factorise $4x^3-7x-3$. Is this answer acceptable?

$(x+\frac{1}{2})(x-\frac{3}{2})(x+1)$.

• Can you add more details to your question, like why do you think your answer might not be acceptable and etc. Asking only , is it correct? Kinda doesn't seem right. You can get anybody to check it. – Mann Apr 16 '15 at 12:45
• Because I don't know is it must to put (2x+1)(2x-3)(x+1). – Mathxx Apr 16 '15 at 12:46
• Yes, add this in your question. – Mann Apr 16 '15 at 12:48

Not quite. What you've got is a quarter of what you want, the correct expression is $$4(x+\frac{1}{2})(x-\frac{3}{2})(x+1)$$ You've done the hard bit - finding the factors - and for most things (most importantly finding roots) your expression is perfectly fine as the four doesn't have much effect. However, if I'm being picky, or if you're being marked on this, then having the four there to have the exact same expression is important.

No. If you expand $(x+\frac{1}{2})\,(x−\frac{3}{2})\,(x+1)$ you get $\displaystyle\frac{4x^3 - 7x - 3}{4}$. You are missing a factor $4$. You can express it like this:

$4x^3 - 7x - 3 = 4\,(x+\frac{1}{2})\,(x−\frac{3}{2})\,(x+1)$.

If you want to avoid the fractions you can also express it as

$(2x+1)\,(2x−3)\,(x+1)$.

First, it is clear that $-1$ is a root. So, $(x+1)$ is a factor of that polynomial. Thus,

$4x^3-7x-3=(x+1)(ax^2+bx+c)=ax^3+bx^2+cx+ax^2+bx+c$

By equating the coefficients of powers, we get:

$a=4$

$b+a=0$ so $b=-4$

$c+b=-7$, so $c=-3$

So, we have $(x+1)(4x^2-4x-3)=4(x+1)(x^2-x-\frac{3}{4})$

Now, using the "usual" method to find roots of quadratic equations, we find that $-\frac{1}{2},\frac{3}{2}$ are roots and $x^2-x-\frac{3}{4}$ is a monic polynomial so it can be written as

$(x+\frac{1}{2})(x-\frac{3}{2})$

Thus, the correct factorization is:

$4(x+1)(x+\frac{1}{2})(x-\frac{3}{2})$

Yes! That is the correct answer if multiplied by 4.

Yes you are almost correct, the correct factor is

$(2x+1)(2x-3)(x+1)$

• Well it depends on, if you are equating to cubic to 0, or considering it of the form $f(x)=somecubic$ , in the latter case. Not putting a 4 in front can cause trouble. – Mann Apr 16 '15 at 12:57