Or: how to avoid Polynomial Long Division when finding factors

Example 6 – Factoring a Polynomial: Repeated Division. Show that (x – 2) and (x + 3) are factors of 2f. (x) = 2x4 + 7x3 – 4x – 27x – 18. The Factor Theorem & Synthetic Substitution. Add to Favorites. 3 teachers like this lesson. Print Lesson. Converse, and I'm not sure it's worth the time, so we'll assume that it's also true that if f(b) = 0, then (x - b) is a factor of the polynomial. Synthetic Substitution Notes.pdf. Student Solutions. • Use long division to divide polynomials by other polynomials. • Use synthetic division to divide polynomials by binomials of the form (x – k). • Use the Remainder Theorem and the Factor Theorem. What You Should Learn.

Do you remember doing division in Arithmetic?

'7 divided by 2 equals 3 with a remainder of 1'

Each part of the division has names:

Which can be rewritten as a sum like this:

Polynomials

Well, we can also divide polynomials.

f(x) ÷ d(x) = q(x) with a remainder of r(x)

But it is better to write it as a sum like this:

Like in this example using Polynomial Long Division:

Example: 2x²−5x−1 divided by x−3

• f(x) is 2x²−5x−1
• d(x) is x−3

After dividing we get the answer 2x+1, but there is a remainder of 2.

• q(x) is 2x+1
• r(x) is 2

In the style f(x) = d(x)·q(x) + r(x) we can write:

2x²−5x−1 = (x−3)(2x+1) + 2

But you need to know one more thing:

Say we divide by a polynomial of degree 1 (such as 'x−3') the remainder will have degree 0 (in other words a constant, like '4').

We will use that idea in the 'Remainder Theorem':

The Remainder Theorem

When we divide f(x) by the simple polynomial x−c we get:

f(x) = (x−c)·q(x) + r(x)

x−c is degree 1, so r(x) must have degree 0, so it is just some constant r:

f(x) = (x−c)·q(x) + r

Now see what happens when we have x equal to c:

So we get this:

The Remainder Theorem:

When we divide a polynomial f(x) by x−c the remainder is f(c)

So to find the remainder after dividing by x-c we don't need to do any division:

Just calculate f(c).

Let us see that in practice:

Example: The remainder after 2x²−5x−1 is divided by x−3

(Our example from above)

We don't need to divide by (x−3) ... just calculate f(3):

2(3)²−5(3)−1 = 2x9−5x3−1
= 18−15−1
= 2

And that is the remainder we got from our calculations above.

We didn't need to do Long Division at all!

Example: The remainder after 2x²−5x−1 is divided by x−5

Same example as above but this time we divide by 'x−5'

'c' is 5, so let us check f(5):

2(5)²−5(5)−1 = 2x25−5x5−1
= 50−25−1
= 24

The remainder is 24

Once again ... We didn't need to do Long Division to find that.

Polynomial Factor Theorem Pdf Calculator

The Factor Theorem

Now ...

What if we calculate f(c) and it is 0?

... that means the remainder is 0, and ...

... (x−c) must be a factor of the polynomial!

We see this when dividing whole numbers. For example 60 ÷ 20 = 3 with no remainder. So 20 must be a factor of 60.

Example: x²−3x−4

f(4) = (4)²−3(4)−4 = 16−12−4 = 0

so (x−4) must be a factor of x²−3x−4

And so we have:

The Factor Theorem:

When f(c)=0 then x−c is a factor of f(x)

And the other way around, too:

When x−c is a factor of f(x) then f(c)=0

Why Is This Useful?

Knowing that x−c is a factor is the same as knowing that c is a root (and vice versa).

The factor 'x−c' and the root 'c' are the same thing

Know one and we know the other

For one thing, it means that we can quickly check if (x−c) is a factor of the polynomial.

Example: Find the factors of 2x³−x²−7x+2

The polynomial is degree 3, and could be difficult to solve. So let us plot it first:

The curve crosses the x-axis at three points, and one of them might be at 2. We can check easily:

Division Theorem For Polynomials

f(2) = 2(2)³−(2)²−7(2)+2
= 16−4−14+2
= 0

Yes! f(2)=0, so we have found a root and a factor.

How about where it crosses near −1.8?

f(−1.8) = 2(−1.8)³−(−1.8)²−7(−1.8)+2
= −11.664−3.24+12.6+2
= −0.304

No, (x+1.8) is not a factor. We could try some other values near by and maybe get lucky.

But at least we know (x−2) is a factor, so let's use Polynomial Long Division:

As expected the remainder is zero.

Better still, we are left with the quadratic equation2x²+3x−1 which is easy to solve.

It's roots are −1.78... and 0.28..., so the final result is:

2x³−x²−7x+2 = (x−2)(x+1.78...)(x−0.28...)

We were able to solve a difficult polynomial.

Summary

The Remainder Theorem:

• When we divide a polynomial f(x) by x−c the remainder is f(c)

The Factor Theorem:

• When f(c)=0 then x−c is a factor of f(x)
• When x−c is a factor of f(x) then f(c)=0

In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is a special case of the polynomial remainder theorem.^[1]

The factor theorem states that a polynomial ${\displaystyle f(x)}$ has a factor ${\displaystyle (x-k)}$if and only if${\displaystyle f(k)=0}$ (i.e. ${\displaystyle k}$ is a root).^[2]

Factorization of polynomials[edit]

Two problems where the factor theorem is commonly applied are those of factoring a polynomial and finding the roots of a polynomial equation; it is a direct consequence of the theorem that these problems are essentially equivalent.

The factor theorem is also used to remove known zeros from a polynomial while leaving all unknown zeros intact, thus producing a lower degree polynomial whose zeros may be easier to find. Abstractly, the method is as follows:^[3]

1. 'Guess' a zero ${\displaystyle a}$ of the polynomial ${\displaystyle f}$. (In general, this can be very hard, but maths textbook problems that involve solving a polynomial equation are often designed so that some roots are easy to discover.)
2. Use the factor theorem to conclude that ${\displaystyle (x-a)}$ is a factor of ${\displaystyle f(x)}$.
3. Compute the polynomial ${\displaystyle g(x)=f(x)/(x-a)}$, for example using polynomial long division or synthetic division.
4. Conclude that any root ${\displaystyle x\ne a}$ of ${\displaystyle f(x)=0}$ is a root of ${\displaystyle g(x)=0}$. Since the polynomial degree of ${\displaystyle g}$ is one less than that of ${\displaystyle f}$, it is 'simpler' to find the remaining zeros by studying ${\displaystyle g}$.

Example[edit]

Find the factors of

${\displaystyle x^3+7x^2+8x+2.}$

To do this one would use trial and error (or the rational root theorem) to find the first x value that causes the expression to equal zero. To find out if ${\displaystyle (x-1)}$ is a factor, substitute ${\displaystyle x=1}$ into the polynomial above:

${\displaystyle x^3+7x^2+8x+2=(1{)}^3+7(1{)}^2+8(1)+2}$

${\displaystyle =1+7+8+2}$

${\displaystyle =18.}$

As this is equal to 18 and not 0 this means ${\displaystyle (x-1)}$ is not a factor of ${\displaystyle x^3+7x^2+8x+2}$. So, we next try ${\displaystyle (x+1)}$ (substituting ${\displaystyle x=-1}$ into the polynomial):

${\displaystyle (-1{)}^3+7(-1{)}^2+8(-1)+2.}$

This is equal to ${\displaystyle 0}$. Therefore ${\displaystyle x-(-1)}$, which is to say ${\displaystyle x+1}$, is a factor, and ${\displaystyle -1}$ is a root of ${\displaystyle x^3+7x^2+8x+2.}$

The next two roots can be found by algebraically dividing ${\displaystyle x^3+7x^2+8x+2}$ by ${\displaystyle (x+1)}$ to get a quadratic:

${\displaystyle \frac{x^3+7x^2+8x+2}{x+1}=x^2+6x+2,}$

and therefore ${\displaystyle (x+1)}$ and ${\displaystyle x^2+6x+2}$ are factors of ${\displaystyle x^3+7x^2+8x+2.}$ Of these the quadratic factor can be further factored using the quadratic formula, which gives as roots of the quadratic ${\displaystyle -3\pm \sqrt{7}.}$ Thus the three irreducible factors of the original polynomial are ${\displaystyle x+1,}$${\displaystyle x-(-3+\sqrt{7}),}$ and ${\displaystyle x-(-3-\sqrt{7}).}$

References[edit]

1. ^Sullivan, Michael (1996), Algebra and Trigonometry, Prentice Hall, p. 381, ISBN0-13-370149-2.
2. ^Sehgal, V K; Gupta, Sonal, Longman ICSE Mathematics Class 10, Dorling Kindersley (India), p. 119, ISBN978-81-317-2816-1.
3. ^Bansal, R. K., Comprehensive Mathematics IX, Laxmi Publications, p. 142, ISBN81-7008-629-9.