Cubic Equation. (Factorisation)

Question

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

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

Answer 1

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

Answer 2

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

$(2x+1)\,(2x−3)\,(x+1)$.

Answer 3

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.

Answer 4

Yes you are almost correct, the correct factor is

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

which is equal to your answer multiplied by 4.

Another note, In Abstrat Algebra, factorization highly depends on what field are we considering. But based from the tag of the question I know that we are pertaining in the field of Real Numbers.

Answer 5

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:

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