## 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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