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The Factor Theorem and Algebraic Division
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The Factor Theorem and Algebraic Division

In science you will often use algebra to solve problems. This article describes two important tools in algebra which are the factor theorem and algebraic division.
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© University of Nottingham

In science you will often use algebra to solve problems. This article describes two important tools in algebra which are the factor theorem and algebraic division.

The Factor Theorem

The factor theorem states that for some function (f(x)), if (f(a)=0), then ((x-a)) is a factor of (f).

An example
a) Find the factors of (f(x)=x^{2}-5x+6)
b) Hence solve (x^{2}-5x+6=0)
a) If we replace (x) with a number, for example, (x=3), then this becomes:

[f(3)=3^{2}-5times 3+6] [f(3)=0]

Therefore the factor theorem states that ((x-3)) must be a factor.
We also find that

[f(2)=2^{2}-5times 2 +6]

(f(2)=0).
Therefore ((x-2)) is also a factor by the factor theorem.
We can thus write that

[f(x)=x^{2}-5x+6=(x-3)(x-2)]

b) Using our factors above we get (f(x)=(x-3)(x-2)=0) which gives us (x=3) or (x=2.)

Algebraic Division

We are now going to look at how we can divide one polynomial by another polynomial. The first method is to use the bus stop method.
An example Divide (x^{3}+2x^{2}-17x+6) by ((x-3)).
Step 1: divide the first term in the bus stop by the first term outside the division sign, and write it on top. For example, (x^{3}div x = x^{2})

[begin{align*} &text{ }text{ }text{ }x^{2} x-3 &overline{big)x^{3}+2x^{2}-17x+6} end{align*}]

Step 2: multiply the first term on top of the bus stop by the expression outside of the division sign, and write it below. For example, (x^{2}times x = x^{3}) and (x^{2}times-3=-3x^{2})

[begin{align*} &text{ }text{ }text{ }x^{2} x-3 &overline{big)x^{3}+2x^{2}-17x+6} &text{ }x^{3}-3x^{2} end{align*}]

Step 3: subtract the first 2 terms just below the bus stop from the first 2 terms in the bus stop. For example, (x^{3}-x^{3}=0) and (2x^{2}−−3x^{2}=5x^{2})

[begin{align*} &text{ }text{ }text{ }x^{2} x-3 &overline{big)x^{3}+2x^{2}-17x+6} &underline{-(x^{3}-3x^{2})} &text{ }text{ }text{ }0+5x^{2} end{align*}]

Step 4: repeat – divide the first term after the subtraction by the first term outside the division sign, and write it on top. For example, (5x^{2}div x =5x)

[begin{align*} &text{ }text{ }text{ }x^{2}+5x x-3 &overline{big)x^{3}+2x^{2}-17x+6} &underline{-(x^{3}-3x^{2})} &text{ }text{ }text{ }0+5x^{2} end{align*}]

Step 5: repeat – multiply the second term on top of the bus stop by the expression out side of the division sign, and write it below. For example, (5xtimes x=5x^{2}) and (5xtimes −3= −15x)

[begin{align*} &text{ }text{ }text{ }x^{2}+5x x-3 &overline{big)x^{3}+2x^{2}-17x+6} &underline{-(x^{3}-3x^{2})} &text{ }text{ }text{ }0+5x^{2} &text{ }text{ }underline{text{ }5x^{2}-15x} end{align*}]

Step 6: repeat – subtract the terms below the bus stop from the corresponding terms above.
For example, (5x^{2}−5x^{2}=0) and (−17x−−15x=−2x)

[begin{align*} &text{ }text{ }text{ }x^{2}+5x x-3 &overline{big)x^{3}+2x^{2}-17x+6} &underline{-(x^{3}-3x^{2})} &text{ }text{ }text{ }0+5x^{2} &text{ }text{ }underline{-(5x^{2}-15x)} &text{ }text{ }text{ }text{ }text{ }text{ }0-2x end{align*}]

Step 7: repeat – divide the first term after the subtraction by the first term outside the division sign, and write it on top. For example, (−2x div x =−2)

[begin{align*} &text{ }text{ }text{ }x^{2}+5x-2 x-3 &overline{big)x^{3}+2x^{2}-17x+6} &underline{-(x^{3}-3x^{2})} &text{ }text{ }text{ }0+5x^{2} &text{ }text{ }underline{-(5x^{2}-15x)} &text{ }text{ }text{ }text{ }text{ }text{ }0-2x end{align*}]

Step 8: repeat – multiply the third term on top of the bus stop by the expression outside of the division sign, and write it below. For example, (−2times x=−2x) and (−2times −3=6)

[begin{align*} &text{ }text{ }text{ }x^{2}+5x-2 x-3 &overline{big)x^{3}+2x^{2}-17x+6} &underline{-(x^{3}-3x^{2})} &text{ }text{ }text{ }0+5x^{2} &text{ }text{ }underline{-(5x^{2}-15x)} &text{ }text{ }text{ }text{ }text{ }text{ }0-2x &text{ }text{ }text{ }text{ }text{ }text{ }-2x+6 end{align*}]

Step 9: repeat – subtract the terms below the bus stop from the corresponding terms above.
For example, (−2x−−2x=0) and (6−6=0)

[begin{align*} &text{ }text{ }text{ }x^{2}+5x-2 x-3 &overline{big)x^{3}+2x^{2}-17x+6} &underline{-(x^{3}-3x^{2})} &text{ }text{ }text{ }0+5x^{2} &text{ }text{ }underline{-(5x^{2}-15x)} &text{ }text{ }text{ }text{ }text{ }text{ }0-2x &text{ }text{ }text{ }text{ }text{ }underline{text{ }-(-2x+6)} &text{ }text{ }text{ }text{ }text{ }text{ }text{ }0+0 end{align*}]

Therefore, ((x^{3}+2x^{2}-17x+6)div (x-3)=x^{2}+5x-2.)
Alternatively we may solve the problem above using algebra.
An example Divide (x^{3}+2x^{2}-17x+6) by ((x-3)).
We may write the problem as solving

[x^{3}+2x^{2}-17x+6=(x-3)(ax^{2}+bx+c)]

Using multiplication we find

[x^{3}+2x^{2}-17x+6=ax^{3}-3ax^{2}+bx^{2}-3bx+cx-3c]

By collecting like terms (x^{3}=ax^{3}), (2x^{2}=-3ax^{2}+bx^{2}), (-17x=-3bx+cx) and (6=-3c).
Clearly (a=1) and we can also find (b) using

[begin{eqnarray*} 2x^{2}&=&-3ax^{2}+bx^{2} 2&=&-3a+b 2&=&-3+b text{ (Using }a=1text{)}. b&=&5 end{eqnarray*}]

Finally (6=-3c) so that (c=-2).
Thus we may write

[x^{3}+2x^{2}-17x+6=(x-3)(x^{2}+5x+-2)]

and thus (frac{x^{3}+2x^{2}-17x+6}{x-3}=x^{2}+5x-2).

Exercise:

Divide (2x^{3}+x^{2}+x-1) by (x-1).

a) What is different about this problem?

b) What is the solution?

© University of Nottingham
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