Polynomials

Hong Kong

Stage 4 - Stage 5

Lesson

In algebra, the remainder theorem allows us to divide one polynomial by another polynomial of the same or lower degree, using a process similar to long division. You can do it without a calculator because it breaks a complex-looking problem up into smaller ones that you have learnt how to solve previously.

The Remainder Theorem is a useful mathematical theorem that can be used to factorize polynomials of any degree in a neat and fast manner.

Solve: $\left(3x^3+2x^2-6\right)\div\left(x+2\right)$(3`x`3+2`x`2−6)÷(`x`+2) using long division

The dividend: $3x^3+2x^2-6$3`x`3+2`x`2−6

The divisor: $x+2$`x`+2

It's helpful to write the dividend like this: $3x^3+2x^2+0x-6$3`x`3+2`x`2+0`x`−6.

Then we can use long division to solve the problem.

1. Divide the first term of the dividend by the highest term of the divisor (meaning the one with the highest power of $x$`x`, which in this case is $x$`x`). Place the result above the bar ($3x^3\div x=3x^2$3`x`3÷`x`=3`x`2).

2. Multiply the divisor by the number you just wrote above the top line. Write the result under the first two terms of the dividend. ($3x^2\times\left(x+2\right)=3x^3+6x^2$3`x`2×(`x`+2)=3`x`3+6`x`2).

3. Subtract these terms from the from those in the original dividend, making sure you pay attention to the positive and negative signs ($3x^3+2x^2-\left(3x^3+6x^2\right)=-4x^2+0x$3`x`3+2`x`2−(3`x`3+6`x`2)=−4`x`2+0`x`).

4. Repeat the previous three steps, except this time use the two terms that have just been written as the dividend (ie. divide $-4x^2+0x$−4`x`2+0`x` by $x+2$`x`+2)

.

5. Repeat step 4. Keep going until there is nothing to "pull down".

The polynomial above the bar is the quotient, $Q(x)$`Q`(`x`), and the number left over, $-22$−22, is the remainder, $R(x)$`R`(`x`).

The remainder theorem also points out the relationship between division and multiplication. For example, since $20\div5=4$20÷5=4, we can also say $4\times5=20$4×5=20. If you get a remainder, then you do the multiplication first, then add the remainder back in. For example, $17\div2=8$17÷2=8 remainder $1$1, which we could also write as $2\times8+1=17$2×8+1=17. The process is the same for polynomials.

The remainder theorem states that when a polynomial ($P(x)$`P`(`x`)) is divided by a linear factor ($x-a$`x`−`a`), the result will be a quotient polynomial ($Q(x)$`Q`(`x`)) plus a remainder ($R(x)$`R`(`x`)). In other words:

$P(x)\div\left(x-a\right)=Q(x)+R(x)$`P`(`x`)÷(`x`−`a`)=`Q`(`x`)+`R`(`x`)

This also means that: $P(x)=\left(x-a\right)\times Q(x)+R(x)$`P`(`x`)=(`x`−`a`)×`Q`(`x`)+`R`(`x`)

You can't always easily solve questions with this quick method but it's handy to know anyway.

Finding remainders

If a polynomial $P(x)$`P`(`x`) is divided by $x-a$`x`−`a`, then: $P(a)=remainder$`P`(`a`)=`r``e``m``a``i``n``d``e``r`

Remember

If the quotient of a polynomial does not leave a remainder, then this value is a factor.

When $3x^3-2x^2-4x+k$3`x`3−2`x`2−4`x`+`k` is divided by $x-3$`x`−3, the remainder is $47$47. Find the value of $k$`k`.

Consider the division $\left(x^2-6x+1\right)\div\left(x+3\right)$(`x`2−6`x`+1)÷(`x`+3).

What term needs to be brought down to move onto the next step in the algorithm?

$x$ `x`$x$ `x`$+$+ $3$3 $x^2$ `x`2$-$− $6x$6 `x`$+$+ $1$1 $x^2$ `x`2$+$+ $3x$3 `x`$-$− $9x$9 `x`What is the remainder of this division?

$x$ `x`$-$− $9$9 $x$ `x`$+$+ $3$3 $x^2$ `x`2$-$− $6x$6 `x`$+$+ $1$1 $x^2$ `x`2$+$+ $3x$3 `x`$-$− $9x$9 `x`$+$+ $1$1 $-$− $9x$9 `x`$-$− $27$27 Without including the remainder, what is the quotient of this division?

Rewrite $x^2-6x+1$

`x`2−6`x`+1 in terms of the divisor, the quotient and the remainder.$x^2-6x+1$

`x`2−6`x`+1$=$=$\left(x+\editable{}\right)\left(x-\editable{}\right)+\editable{}$(`x`+)(`x`−)+

Consider the following division: $\frac{x^2+17x+70}{x+10}$`x`2+17`x`+70`x`+10.

Perform the division using long division.

$\editable{}$ $+$+ $\editable{}$ $x$ `x`$+$+ $10$10 $x^2$ `x`2$+$+ $17x$17 `x`$+$+ $70$70 $-$− $\editable{}$ $+$+ $\editable{}x$ `x`$\editable{}x$ `x`$+$+ $\editable{}$ $-$− $\editable{}x$ `x`$+$+ $\editable{}$ $\editable{}$ Is $x+10$

`x`+10 a factor of the polynomial $x^2+17x+70$`x`2+17`x`+70?Yes

ANo

BWhat is the other factor?

Consider the following division:

$\left(3x^3-15x^2+2x-10\right)\div\left(x-5\right)$(3`x`3−15`x`2+2`x`−10)÷(`x`−5)

Fill in the gaps to complete the long division process below.

$\editable{}$ $+$+ $0x+$0 `x`+$\editable{}$ $x-5$ `x`−5$3x^3$3 `x`3$-$−$15x^2$15 `x`2$+$+$2x$2 `x`$-$−$10$10 $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ State the quotient and remainder when $3x^3-15x^2+2x-10$3

`x`3−15`x`2+2`x`−10 is divided by $x-5$`x`−5.Quotient = $\editable{}$

Remainder = $\editable{}$