remainder theorem , factor theorem , rational root theorem ♡

remainder theorem

The remainder theorem states when polynomial $f (x)$ is divided by $(a x + b)$ then the remainder is equal to $f (\frac{- b}{a})$ .

Example questions.

Find the remainder when $f (x) = x^{3} - 3 x^{2} + 2 x + 6$ is divided by $x - 2$ .

$f (2) = 6$

Find the remainder when $f (x) = x^{3} - 2 x + 4$ is divided by $2 x + 1$ .

$f (\frac{- 1}{2}) = \frac{39}{8}$

factor theorem

The factor theorem states when polynomial $f (x)$ is divided by $(a x + b)$ and the remainder is zero, then $(a x + b)$ is a factor of $f$ .

Using the remainder theorem as well, we have the statement:

For a polynomial $f (x)$ , the statements ” $a x + b$ is a factor” and " $f (\frac{- b}{a}) = 0$ " are equivalent.

It’s pretty intuitive yes?

Applications of the factor theorem:

evaluating to see if $(a x + b)$ is a factor of $f (x)$
finding coefficients of $f (x)$ when given the factor or remainder of $f (x)$ by $(a x + b)$

Given the polynomial $f (x) = 3 x^{2} - 4 x + 7$ , determine if $(x - 1)$ is a factor.

If the remainder of $f$ divided by $(x - 1)$ is equal to $f (1)$ then $(x - 1)$ is a factor of $f$ .

When we divide $f$ by $(x - 1)$ we get $3 x - 1$ with a remainder of $6$ , and $f (1) = 6$ . The answer is yes.

Show that $x + 1$ is a factor of $x^{3} - 4 x^{2} + x + 6$ and hence find other factors.

$f (- 1) = 0$ means that $x + 1$ is a factor.

Perform division to factorise.

$= (x + 1) (x^{2} - 5 x + 6)$
$= (x + 1) (x - 3) (x - 2)$

Factorise $x^{3} - 2 x^{2} - 5 x + 6$ .

Step one, trial and error.

$f (- 1) = 8$
$f (1) = 0$ means that $x - 1$ is a factor.

Perform division to factorise.

$= (x - 1) (x^{2} - x - 6)$
$= (x - 1) (x - 3) (x + 2)$

rational root theorem

The rational root theorem helps find possible rational roots (x-intercepts) of a polynomial equation. It states that if a polynomial has integer coefficients, any rational root, expressed as a fraction p/q, must satisfy: p is a factor of the constant term, and q is a factor of the leading coefficient.

Now action.

For a polynomial $f (x)$ , all of the possible rational roots (x-intercepts) of the function reside within $x = \pm \frac{any one factor of constant term}{any one factor of leading coefficient}$

Why is the above statement true? Umm here.

Remember to try the factors that are closer to zero first, since they are small.

After you get a possible root, divide your polynomial using synthetic division instead of polynomial long division, cause easier.

examples

worked example: factor $x^{3} - 2 x^{2} - 5 x + 6$
$x \in {1, - 1, 2, - 2, 3, - 3, 6, - 6}$
try $(x - 1)$
$(x - 1) (x^{2} - x - 6)$
$(x - 1) (x + 2) (x - 3)$

worked example: factor $2 x^{3} + x^{2} - 5 x + 2$
$x \in {1, - 1, 2, - 2}$
try $(x - 1)$

worked example: factor $x^{3} + 3 x^{2} - 6 x - 8$
$x \in {1, - 1, 2, - 2, 4, - 4, 8, - 8}$
try $(x - 1)$
we get a remainder
try $(x + 1)$
$(x + 1) (x^{2} + 2 x - 8)$
$(x + 1) (x - 2) (x + 3)$

worked example: factor $x^{3} + 2 x^{2} - 9 x - 18$
$x \in {1, - 1, 2, - 2, 3, - 3, 6, - 6 \dots}$
try $(x - 1)$
we get a remainder
try $(x + 1)$
we get a remainder
try $(x - 2)$
we get a remainder
try $(x + 2)$
no remainder
$(x + 2) (x^{2} - 9)$
$(x + 2) (x + 3) (x - 3)$
x

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remainder theorem , factor theorem , rational root theorem