## The Astounding Power of Area

### 2.6 The Remainder Theorem/Factor Theorem

Here is something curious.

The reverse area method shows that $$p\left(x\right) = x^{4}-2x^{3}+3x^{2}-7x+9$$ leaves a remainder of $$7$$ when divided by $$x-2$$.

(We have $$\dfrac{x^{4}-2x^{3}+3x^{2}-7x+9}{x-2}=x^{3}+3x-1+\dfrac{7}{x-2}$$.  Check this!)

And the value of $$p\left(2\right)$$ is $$2^{4}-2\cdot2^{3}+3\cdot2^{2}-7\cdot2+9=7$$.

This is not a coincidence.

1. The reverse area method shows that dividing a polynomial $$p\left(x\right)$$ by a linear term $$x-h$$ gives a remainder (if any) that is just a single number $$a$$ (no $$x$$ terms): $$\dfrac{p\left(x\right)}{x-h}=q\left(x\right) + \dfrac{a}{x-h}$$ for some other polynomial $$q$$.

(If, for any reason, we have $$\dfrac{p\left(x\right)}{x-h}=q\left(x\right) + \dfrac{ax^{2}+bx+c}{x-h}$$, say, or $$\dfrac{p\left(x\right)}{x-h}=q\left(x\right) + \dfrac{ax+b}{x-h}$$, we can perform yet another division on the ratio term starting at the top left cell to reduce the power of $$x$$ that appears in a remainder.)

2. Multiplying through by $$x-h$$ gives $$p\left(x\right) = q\left(x\right)\left(x-h\right) + a$$.

3. Putting $$x=h$$ shows $$p\left(h\right) = a$$. That is, the remainder $$a$$ is the value of the polynomial at $$x=h$$.

THE REMAINDER THEOREM

When dividing a polynomial $$p\left(x\right)$$ by a linear term $$x-h$$, the remainder is $$p\left(h\right)$$. We have $$\dfrac{p\left(x\right)}{x-h}=q\left(x\right) + \dfrac{p\left(h\right)}{x-h}$$.

Taking it a bit further, we see $$p\left(x\right) = q\left(x\right)\left(x-h\right) + p\left(h\right)$$.

It follows that if $$p\left(h\right) = 0$$, then $$x-h$$ is a factor of the polynomial.

THE FACTOR THEOREM

If $$x=h$$ is a zero of a polynomial $$p\left(x\right)$$, that is, $$p\left(h\right)=0$$, then $$x-h$$ is a factor of the polynomial.

(And, conversely, if $$x-h$$ is a factor of a polynomial, that is, $$p\left(x\right) = \left(x-h\right)q\left(x\right)$$, then $$p\left(h\right)=0$$.

EXERCISE:  Quickly show that $$p\left(x\right) = x^{5}+ x^{4}-7x^{3}-x^{2}+6x$$ has factors $$x-1$$, $$x+1$$, $$x-2$$, and $$x$$. Now factor $$p\left(x\right)$$ completely.

AN ALTERNATIVE SLICK WAY TO THE FACTOR THEOREM

We saw in lesson 2.3 that $$x^{n}-a^{n}$$ is evenly divisible by $$x-a$$. So if $$p\left(x\right) = b_{n}x^{n}+b_{n-1}x^{n-1}+\ldots+b_{1}x+b_{0}$$, then $$p\left(x\right) – p\left(a\right)$$ is evenly divisible by $$x-a$$ too.

$$\dfrac{ p\left(x\right) – p\left(a\right)}{x-a}=q\left(x\right)$$ for some polynomial $$q$$.

If it turns out that $$p\left(a\right) = 0$$, then this reads

$$\dfrac{p\left(x\right)}{x-a}=q\left(x\right)$$

and we see that $$x-a$$ is a factor of $$p\left(x\right)$$.

EXERCISE SOLUTION:

Check that $$p\left(1\right)=0$$, $$p\left(-1\right)=0$$, $$p\left(2\right)=0$$, and $$p\left(0\right)=0$$. This shows that $$x-1$$, $$x+1$$, $$x-2$$, and $$x-0=x$$ are each factors of the polynomial. So the polynomial is evenly divisible by $$\left(x-1\right)\left(x+1\right)\left(x-2\right)x=x^{4}-2x^{3}-x^{2}+2x$$. Dividing $$p\left(x\right)$$ by this polynomial shows that the remaining factor is $$x+3$$. So $$p\left(x\right)=x\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+3\right)$$.

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