4 Quadratics and polynomials

Lesson

Polynomials can be added, subtracted and multiplied together. They can also be divided.

We're familiar with the idea of factorising quadratic polynomials. This idea can be extended to polynomials of any degree. That is, a polynomial of any degree can be rewritten as a product of polynomials of a lesser degree.

Quadratic polynomials, can sometimes be factorised into the product of two linear polynomials. That is, a polynomial of degree $2$2 can be factorised into two polynomials of degree $1$1.

For a polynomial of any degree, the sum of the degrees of the factors will equal the degree of the original polynomial. For example, a polynomial of degree $4$4 can be factorised into $4$4 polynomials of degree $1$1 or $2$2 polynomials of degree $2$2.

If $A\left(x\right)$`A`(`x`) is a factor of $P\left(x\right)$`P`(`x`) then that means that $A\left(x\right)$`A`(`x`) evenly divides $P\left(x\right)$`P`(`x`). If $P\left(x\right)=A\left(x\right)Q\left(x\right)$`P`(`x`)=`A`(`x`)`Q`(`x`) then it follows that $\frac{P\left(x\right)}{A\left(x\right)}=Q\left(x\right)$`P`(`x`)`A`(`x`)=`Q`(`x`). This is how we can divide polynomials.

Not all polynomials can be evenly divided by other polynomials, in the same way that not all integers can be evenly divided by other integers. However, in the case where they do not evenly divide, there will be a remainder which will also be a polynomial.

In general, for any polynomials $P\left(x\right)$`P`(`x`) and $A\left(x\right)$`A`(`x`) there will be polynomials $Q\left(x\right)$`Q`(`x`) and $R\left(x\right)$`R`(`x`) such that $P\left(x\right)=A\left(x\right)Q\left(x\right)+R\left(x\right)$`P`(`x`)=`A`(`x`)`Q`(`x`)+`R`(`x`). The sum of the degrees of $A\left(x\right)$`A`(`x`) and $Q\left(x\right)$`Q`(`x`) will equal the degree of $P\left(x\right)$`P`(`x`) and the degree of $R\left(x\right)$`R`(`x`) will be less than the degree of $A\left(x\right)$`A`(`x`).

If $A\left(x\right)=x-a$`A`(`x`)=`x`−`a` for some number $a$`a` then $R\left(x\right)=P\left(a\right)$`R`(`x`)=`P`(`a`). In other words, the remainder will be the value of $P\left(x\right)$`P`(`x`) with $a$`a` substituted for $x$`x`. This is called the remainder theorem.

If $A\left(x\right)$`A`(`x`) is a factor of $P\left(x\right)$`P`(`x`) then the remainder $R\left(x\right)=0$`R`(`x`)=0. From the remainder theorem, this means that if $A\left(x\right)=x-a$`A`(`x`)=`x`−`a` then $P\left(a\right)=0$`P`(`a`)=0. The converse is also true, so if $P\left(a\right)=0$`P`(`a`)=0 then $x-a$`x`−`a` is a factor of $P\left(x\right)$`P`(`x`). This is called the factor theorem. We call values of $a$`a` where $P\left(a\right)=0$`P`(`a`)=0 zeros of the polynomial $P\left(x\right)$`P`(`x`).

A method we can use to divide polynomials is called polynomial long division. This works similarly to long division of numbers. Watch this video to learn how to do polynomial long division.

Summary

For polynomials $P\left(x\right)$`P`(`x`) and $A\left(x\right)$`A`(`x`) there are polynomials $Q\left(x\right)$`Q`(`x`) and $R\left(x\right)$`R`(`x`) such that $P\left(x\right)=A\left(x\right)Q\left(x\right)+R\left(x\right)$`P`(`x`)=`A`(`x`)`Q`(`x`)+`R`(`x`). In this equation:

- $A\left(x\right)$
`A`(`x`) is the divisor - $Q\left(x\right)$
`Q`(`x`) is the quotient - $R\left(x\right)$
`R`(`x`) is the remainder

The remainder theorem says that if $A\left(x\right)=x-a$`A`(`x`)=`x`−`a` then $R\left(x\right)=P\left(a\right)$`R`(`x`)=`P`(`a`).

The factor theorem says that if $P\left(a\right)=0$`P`(`a`)=0, then $A\left(x\right)=x-a$`A`(`x`)=`x`−`a` is a factor of $P\left(x\right)$`P`(`x`).

The converse of the factor theorem says that if $A\left(x\right)=x-a$`A`(`x`)=`x`−`a` is a factor of $P\left(x\right)$`P`(`x`) then $P\left(a\right)=0$`P`(`a`)=0.

We can divide polynomials using polynomial long division.

Using the remainder theorem, find the remainder when $P\left(x\right)=3x^4+5x^3-2x^2+6x+7$`P`(`x`)=3`x`4+5`x`3−2`x`2+6`x`+7 is divided by $A\left(x\right)=3x+1$`A`(`x`)=3`x`+1.

Using the factor theorem or otherwise, rewrite $x^3-6x^2+11x-6$`x`3−6`x`2+11`x`−6 as a product of linear factors.

If the distance travelled is $\left(3x^3-2x^2+4x+9\right)$(3`x`3−2`x`2+4`x`+9) kilometres and the speed is $\left(x+1\right)$(`x`+1) km/h, what is the time travelled in hours?

Investigate the concept of a polynomial and apply the factor and remainder theorems to solve problems.