Monomial Factors of Polynomials

Key Questions

The 'normal' way of finding the GCF of two polynomials is to factor both of them completely, then pick out the common factors and multiply them together.

Alternatively you can use division.

Explanation:

Normal Method - Factor both polynomials first

For example:

Given $f \left(x\right) = {x}^{2} + 6 x + 9$ and $g \left(x\right) = {x}^{2} + x - 6$,

you can factor $f \left(x\right) = {\left(x + 3\right)}^{2}$ and $g \left(x\right) = \left(x + 3\right) \left(x - 2\right)$

hence the GCF is $\left(x + 3\right)$

Backup Method - Using division of polynomials

Given: $f \left(x\right) = {x}^{4} + 2 {x}^{3} + 4 {x}^{2} + 3 x + 2$
and: $g \left(x\right) = {x}^{4} + 3 {x}^{3} + 6 {x}^{2} + 5 x + 3$

If $f \left(x\right)$ and $g \left(x\right)$ have a common polynomial factor $p \left(x\right)$ then if we divide $f \left(x\right)$ by $g \left(x\right)$ (or vice versa), the remainder must be a multiple of $p \left(x\right)$.

For example, $g \left(x\right) = f \left(x\right) + \left({x}^{3} + 2 {x}^{2} + 2 x + 1\right)$

Let $h \left(x\right) = {x}^{3} + 2 {x}^{2} + 2 x + 1$

If we now divide $f \left(x\right)$ by $h \left(x\right)$ then any remainder will also be divisible by $p \left(x\right)$:

${x}^{4} + 2 {x}^{3} + 4 {x}^{2} + 3 x + 2$

$= \left({x}^{3} + 2 {x}^{2} + 2 x + 1\right) x + 2 {x}^{2} + 2 x + 2$

$= h \left(x\right) \cdot x + 2 \left({x}^{2} + x + 1\right)$

Next try dividing $f \left(x\right)$ by ${x}^{2} + x + 1$ ...

${x}^{4} + 2 {x}^{3} + 4 {x}^{2} + 3 x + 2 = \left({x}^{2} + x + 2\right) \left({x}^{2} + x + 1\right)$

This time there is no remainder, so $p \left(x\right) = {x}^{2} + x + 1$ is our GCF

For a more complex example, see http://socratic.org/questions/if-you-are-told-that-x-7-3x-5-x-4-4x-2-4x-4-0-has-at-least-one-repeated-root-how

• You can check your factoring by multiplying them all out to see if you get the original expression. If you do, your factoring is correct; otherwise, you might want to try again.

I hope that this was helpful.

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