1.2: Square Root Property, Complete The Square, And The Quadratic ...

Solving Cubic Equations

If multiple roots and complex roots are counted, then the fundamental theorem of algebra implies that every polynomial with one variable will have as many roots as its degree. For example, we expect \(f (x) = x^{3} − 8\) to have three roots. In other words, the equation

\(x^{3}-8=0\)

should have three solutions. To find them one might first think of trying to extract the cube roots just as we did with square roots,

\(\begin{aligned} x^{3}-8 &=0 \\ x^{3} &=8 \\ x &=\sqrt[3]{8} \\ x &=2 \end{aligned}\)

As you can see, this leads to one solution, the real cube root. There should be two others; let’s try to find them.

Example \(\PageIndex{19}\):

Find the set of all roots: \(f(x)=x^{3}-8\).

Solution

Notice that the expression \(x^{3}-8\) is a difference of cubes and recall that \(a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\). Here \(a=x\) and \(b=2\) and we can write

\(\begin{aligned} x^{3}-8 &=0 \\(x-2)\left(x^{2}+2 x+4\right) &=0 \end{aligned}\)

Next apply the zero-product property and set each factor equal to zero. After setting the factors equal to zero we can then solve the resulting equation using the appropriate methods.

\(\begin {array}{l c l} { x-2=0} &\text{or} & x^{2}+2x+4=0 \\ {x=2}& & { \begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-(2) \pm \sqrt{(2)^{2}-4 (1)(4)}}{2(1)} \\ &=\frac{-2 \pm \sqrt{-12}}{2} \\ &=\frac{-2 \pm 2 i \sqrt{3}}{2} \\ &=-1 \pm i \sqrt{3} & \text{Solution Set: } \{2,-1 \pm i \sqrt{3}\} \end{aligned} } \end{array}\)

Using this method, we were able to obtain the set of all three roots \(\{2,-1 \pm i \sqrt{3}\}\), one real and two complex.

Sometimes factoring will produce a quadratic factor that needs to be solved using the quadratic formula or complete the square.

Example \(\PageIndex{21}\):

Solve \( 2x^4+2000x=0\)

Solution:

Factor first. \( 2x^4+2000x= 2x(x^3+1000) = 2x(x+10)(x^2-10x+100)\)

Use the zero factor property and complete the square on: \( \quad 2x(x+10)(x^2-10x+100) = 0 \)

\( \begin{array} {lll} 2x=0 & (x+10) = 0 & (x^2-10x+100) = 0 \\ x=0& x=-10 & x^2-10x +\square = -100+\square \\ && (x-5)^2 = -100+25 \\ && x-5 = \pm \sqrt {-75} \\ && x = 5 \pm 5i\sqrt{3} \\ \end{array} \)

The solution set is \( \{ 0, -10, 5 \pm 5i\sqrt{3} \} \)

When an expression is both a difference of squares and a sum or difference of cubes, if factoring as a difference of squares is done first, a more complete factorization is obtained. For example, if given the equation \( 64x^6-1=0\) to solve, when it is first factored as a difference of squares as \( (8x^3-1) (8x^3+1) \), and then as a difference of cubes, the solutions obtained are \( \pm \frac{1}{2}\) and \( \pm \frac{1}{4}({1 \pm i\sqrt{3} }) \). In contrast, if it is first factored as a difference of cubes as \( (4x^2-1) (16x^4 +4x^2+1) \), the solutions eventually obtained are \( \pm \frac{1}{2}\) and \( \pm \frac{1}{4}\sqrt {-2 \pm 2i\sqrt{3} } \).