iGCSE (2021 Edition)

Lesson

Similar to solving quadratics we have a few methods to choose between when solving cubic equations:

**Solve using algebraic manipulation**- For cubics such as $2x^3+16=0$2`x`3+16=0.**Factorise**- Fully factorising the cubic and then we can then use the null factor law to solve. If $a\times b=0$`a`×`b`=0 then either $a=0$`a`=0 or $b=0$`b`=0.Once we have extracted the important information from a question and formed an equation, we could use technology to solve the equation graphically or algebraically.**Technology**-

When factorising a cubic recall 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.

Solve the equation $x^3=-8$`x`3=−8.

Solve the equation $x^3-49x=0$`x`3−49`x`=0.

State the solutions on the same line, separated by a comma.

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.

There are algebraic and graphical methods that we can use to solve inequalities involving the cubes of unknowns. We'll discuss a couple of examples.

Solve $\left(x-2\right)^3\le-27$(`x`−2)3≤−27.

While this is a straightforward example, try to pay close attention to the strategies involved. The same strategies can be used for much more difficult questions.

Using an algebra approach, we write:

$\left(x-2\right)^3$(x−2)3 |
$<=$<= | $-27$−27 |

$x-2$x−2 |
$<=$<= | $-3$−3 |

$x$x |
$<=$<= | $-1$−1 |

Using a graphical approach we can graph $f\left(x\right)=\left(x-2\right)^3$`f`(`x`)=(`x`−2)3 and $g\left(x\right)=-27$`g`(`x`)=−27 on the same set of axes, and then see for what values of $x$`x` is $f\left(x\right)\le g\left(x\right)$`f`(`x`)≤`g`(`x`). The result, shown graphically is $x\le-1$`x`≤−1.

An alternative approach is to rearrange the inequality so that $\left(x-2\right)^3+27\le0$(`x`−2)3+27≤0, and form the function $h\left(x\right)=\left(x-2\right)^3+27$`h`(`x`)=(`x`−2)3+27. Then see for what $x$`x` is $h\left(x\right)\le0$`h`(`x`)≤0. Again, shown graphically, the same result is found.

Solve the inequality $\left(x+2\right)^3>-64$(`x`+2)3>−64.

Use the graph of the function $y=\left(x-1\right)\left(x+1\right)\left(x-3\right)$`y`=(`x`−1)(`x`+1)(`x`−3), provided below, to solve the inequality $\left(x-1\right)\left(x+1\right)\left(x-3\right)\le0$(`x`−1)(`x`+1)(`x`−3)≤0.

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Select all correct regions.

$1\le x\le3$1≤

`x`≤3A$x\le-1$

`x`≤−1B$x\ge3$

`x`≥3C$-1\le x\le1$−1≤

`x`≤1D$x\ge3$

`x`≥3E

Use the graph of the function $y=2x^3-4x^2-30x$`y`=2`x`3−4`x`2−30`x` and the line $y=-72$`y`=−72 below to solve the inequality $2x^3-4x^2-30x<-72$2`x`3−4`x`2−30`x`<−72.

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