Help in factoring polynomials

1. $x^3 - 13x + 12$

2. $x^5 - 3x^3 - 4x$

3. $x^3 - 6x^2 + 5x + 12$

• Some context would help us better help you. – Git Gud Sep 4 '13 at 21:01
• Hints: $1^3-13\cdot 1+12=0$, $(-1)^2-3\cdot(-1)-4=0$, $(-1)^3-6\cdot (-1)^2+5\cdot(-1)+12 = 0$. – njguliyev Sep 4 '13 at 21:05
• in the third problem do i change the sign? (x^3 - 6x^2) + ( 5x + 12 ) or (x^3 - 6x^2) + ( 5x - 12 ) – Zii Sep 4 '13 at 22:21

Hint:

1. $1^3 - 13\cdot1 + 12=0$

2. $x\cdot\left((\pm2)^4 - 3\cdot(\pm2)^2 - 4\right)=0$

3. $(-1)^3 - 6(-1)^2 + 5(-1) + 12=0$

I will do the first.

1. Find the zeros of the polynomial. Since these are high-order polynomials, you can guess the first zero. For example, a zero of $x^3-13x+12$ is $1$.
2. Perform polynomial long division $\left(x^3 - 13x + 12\right)/\left(x - 1\right)=x^2+x-12$
3. Use the quadratic formula to get the remaining zeros of $x^2+x-12$. They are $-4$ and $3$.
4. The answer is $\left(x-1\right)\left(x-3\right)\left(x+4\right)$.

Hint for question 2. Take out the common factor of x to get x(x^4-3x^2-4), then substitute u=x^2 into the bracketed part of this...

HINT:

Try to find the zeroes of the polynomials. If the zero of the polynomial is $x_0$, then divide the polynomial by $x_0$ and then repeat again.

For the first one:

$x_0 = 1$ is zero of the polynomial so we have:

$$\frac{x^3 - 13x + 12}{x-1} = (x^2 + x -12)$$

So the first polynomial can be written as:

$$x^3 - 13x + 12 = (x-1)(x^2+x-12)$$

The zeroes for the second term are $x_0 = 3$ and $x_0 = -4$. So the expression can be written as:

$$x^3 - 13x + 12 = (x-1)(x-3)(x+4)$$

You can do the other two yourself.