# How do you find the zeroes of #P(x)= x^4-2x^3-7x^2+8x+12#?

##### 3 Answers

Find zeros:

#### Explanation:

Given:

#P(x) = x^4-2x^3-7x^2+8x+12#

**Rational roots theorem**

The rational roots theorem tells us that the rational zeros of

That means that the only possible rational roots are:

#+-1, +-2, +-3, +-4, +-6, +-12#

Trying each in turn, we find:

#P(-1) = 1+2-7-8+12 = 0#

So

We could carry on just trying other possible rational roots from the list, but let's factor out the one we have found...

#x^4-2x^3-7x^2+8x+12 = (x+1)(x^3-3x^2-4x+12)#

Notice that in the remaining cubic factor the ratio of the first and second terms is the same as the ratio of the third and fourth terms.

So this cubic will factor by grouping:

#x^3-3x^2-4x+12 = (x^3-3x^2)-(4x-12)#

#color(white)(x^3-3x^2-4x+12) = x^2(x-3)-4(x-3)#

#color(white)(x^3-3x^2-4x+12) = (x^2-4)(x-3)#

#color(white)(x^3-3x^2-4x+12) = (x^2-2^2)(x-3)#

#color(white)(x^3-3x^2-4x+12) = (x-2)(x+2)(x-3)#

So the other three zeros are

Four zeros are

#### Explanation:

According to Factor Theorem if for

Further if

Hence, here if

It is easier if

If

and if

Therefore, three zeros are

If

#### Explanation:

If we **complete the square** of **last term**

will be

**Rearranging** the **terms** of

Hence, the **zeroes** of