Thread: Require aid for simple factorising question...

1. Require aid for simple factorising question...

Hi.

I've not actually been taught how to factorise the following, and I'm required to do so for upcoming exams.

$x^3 - 4x^2 + x + 6$

The book states a simplified version of the above is the following:

$(x - 3)(x^2 - x - 2)$

Could someone please explain to me how you get from the above equation to the simplified one?

Thanks.

2. If you have a polynomial

$P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+..........+a_{0}$

You can find the solution $x=x_{0}$ in the divisors of $a_{0}$.

If you have found a solution $x=x_{0}$ satisfying $P(x_{0})=0$ then you can write your polynomial expresion like this:

$P(x)=(x-x_{0})*Q(x)$

Where Q(x) is calculated by dividing P(x) at $(x-x_{0})$

In your example your $x_{0}=3$ because 3 is satisfying the condition $P(x_{0})=0 \Rightarrow P(3)=0 \Rightarrow$ your

P(x)=(x-3)*Q(x)

You will find $Q(x)=x^2-x-2$ by dividing $P(x)=x^3-4x^2+x+6$ at $x-3$

Hope it's clear

Have a nice day!

3. If (x-a) is a factor of $x^3- 4x^2+ x+ 6$, then we must have [tex]x^3- 4x^2+ x+ 6= (x- a)(bx^3+ cx+ d)[tex]. That tells you two things. First, whatever a is, it must make $a^3- 4a^2+ a+ 6= 0$ because the "(x-a)" factor become "(a-a)= 0" so the product on the right is 0. Further, while it is theoretically possible to factor using irrational numbers there is no good way to do that so you are really looking for factors using rational and integer numbers. Looking just at the constant terms in [tex]x^3- 4x^2+ x+ 6= (x- a)(bx^3+ cx+ d)[tex] it is easy to see that we must have ad= 6 and so a must be a factor of 6. The integer factors of 6 are $\pm 1$, $\pm 2$, [tex]\pm 3[/math and $\pm 6$. Check those numbers to see that 3 makes the polynomial 0 and so x- 3 really is a factor. You can then divide by x- 3 as Hush_Hush says to find the other factor.

Actually, as you are doing that, you may also find that $(-1)^3- 4(-1)^2+ (-1)+ 6= -6+ 6= 0$ so that (x-(-1))= x+ 1 is also a factor and that $2^3- 4(2^2)+ 2+ 6= 16- 16= 0$ so that x-2 is a third factor. That is, $x^3- 4x^2+ x+ 6= (x- 3)(x^2- x- 2)= (x- 3)(x+ 1)(x- 2)$.

4. Originally Posted by cosmos22
I've not actually been taught how to factorise the following, and I'm required to do so for upcoming exams.

$x^3 - 4x^2 + x + 6$
There is a whole suite of tools used for factoring bigger-than-quadratic polynomials. For a run-down, start with solving polynomials, and then learn how to use those "solving" tools for factoring polynomials. If you aren't familiar with the topics, you might want to start with learning about synthetic division and the Rational Roots Test.

Good luck!