Equations

Lesson

There are $2$2 relatively simple methods that can be used to solve the more basic forms of cubic equations.

The first uses the concepts studied here on the null factor theorem. So if we can factorise a cubic equation fully, then we can solve the cubic by solving each of the individual factored elements.

There are many techniques for factorising cubics, these are covered in the factorisation section.

We can factorise by:

- using the highest common factor
- factor special forms such as the sum and difference of cubes
- identifying a single factor, then using division to establish the remaining quadratic. From here you would employ any of the factorising methods for quadratics.

There is a great benefit to factorising equations in order to solve them, in order to understand why we need to think about ZERO.

The property of $0$0 is very special. The ONLY way two, three or any number of things that are being multiplied can have the answer of $0$0, is if one of those things are $0$0 themselves.

So if I have $2$2 factors, like $a$`a` and $b$`b` , and I multiply them together and they equal $0$0. Then one of those factors $a$`a` or $b$`b` MUST be $0$0. A written solution to a question like this would be similar to the following,

If $a\times b=0$`a`×`b`=0 then $a=0$`a`=0 or $b=0$`b`=0

Solve the equation $x^3-36x=0$`x`3−36`x`=0

$x^3-36x=0$`x`3−36`x`=0 becomes $x\left(x^2-36\right)=0$`x`(`x`2−36)=0

and now we can use the difference of $2$2 squares the factorise further getting

$x\left(x-6\right)\left(x+6\right)=0$`x`(`x`−6)(`x`+6)=0

Remembering that the only way a multiplication can result in $0$0 is if one of the elements is $0$0, then we can say that for $x\left(x-6\right)\left(x+6\right)=0$`x`(`x`−6)(`x`+6)=0, either

$x=0$`x`=0 or

$\left(x-6\right)=0$(`x`−6)=0, so $x=6$`x`=6 or

$\left(x+6\right)=0$(`x`+6)=0, so $x=-6$`x`=−6.

Thus the solution to $x\left(x-6\right)\left(x+6\right)=0$`x`(`x`−6)(`x`+6)=0 is $x=0,\pm6$`x`=0,±6

The second option is to manipulate the equation using rules of algebra to solve. Equations that can be solved like this generally have only an $x^3$`x`3 and constant terms, no $x^2$`x`2 terms and no $x$`x` terms.

This question for example could be solved in this manner.

$125x^3-8$125x3−8 |
$=$= | $0$0 |

$125x^3$125x3 |
$=$= | $8$8 |

$x^3$x3 |
$=$= | $\frac{8}{125}$8125 |

$x$x |
$=$= | $\frac{2}{5}$25 |

Solve the following equation:

$x^3-125=0$`x`3−125=0

Solve the equation $x^3=-8$`x`3=−8.

The cubic $P\left(x\right)=x^3-2x^2-5x+6$`P`(`x`)=`x`3−2`x`2−5`x`+6 has a factor of $x-3$`x`−3.

Solve for the roots of the cubic. If there is more than one root, state the solutions on the same line separated by commas.

Form and use linear, quadratic, and simple trigonometric equations

Apply graphical methods in solving problems

Apply algebraic methods in solving problems