We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.
polynomial equation
A polynomial of degree n is an expression of the type
where n is a positive integer and [latex]{a}_{n},\dots ,{a}_{0}[/latex] are real numbers and [latex]{a}_{n}\ne 0[/latex].
Setting the polynomial equal to zero gives a polynomial equation. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent n.
Solving by factoring depends on the zero-product property which states that if [latex]a \cdot b=0[/latex], then [latex]a=0[/latex] or [latex]b=0[/latex], where [latex]a[/latex] and [latex]b[/latex] are real numbers or algebraic expressions. In other words if the product of two expressions equals zero, then at least one of the expressions must equal zero. We may apply the zero-product property to polynomial equations in the same way we did to quadratic equations.How to: Solve polynomial equations by factoring
Ensure Polynomial is Set to Zero: Begin by rewriting the equation so that all terms are on one side and the equation equals zero. This standard form makes it clear which terms need to be factored.
Factor the Polynomial: Decompose the polynomial into its factorable elements. This might involve:
Taking out the Greatest Common Factor (GCF): Always start by factoring out the highest common factor from all terms.
Using Special Products: Look for patterns like difference of squares, perfect square trinomials, or sum/difference of cubes.
Factoring Trinomials: Decompose middle-term trinomials into binomials.
Grouping: For polynomials with more than three terms, use grouping to find common factors within grouped terms.
Set Each Factor Equal to Zero: Set each factor in the factored equation to zero and solve for the variable.
Solve for the Variable: This involves simple algebraic manipulation to isolate the variable in each equation formed from the factors.
Solve the polynomial by factoring:
[latex]5x^4 = 80x^2[/latex]
[latex]\begin{align*} \text{Original equation} & : & 5x^4 &= 80x^2 \\ \text{Subtract $80x^2$ from both sides} & : & 5x^4 - 80x^2 &= 0 \\ \text{Factor out the common term $5x^2$} & : & 5x^2(x^2 - 16) &= 0 \\ \text{Factor $x^2 - 16$ as a difference of squares} & : & 5x^2(x-4)(x+4) &= 0 \\ \end{align*}[/latex]
Applying the zero-product property:
[latex]\begin{align*} \text{Set first factor $5x^2$ to zero} & : & 5x^2 &= 0 \\ \text{Divide both sides by 5} & : & x^2 &= 0 \\ \text{Take the square root of both sides} & : & x &= 0 \\ \text{Set the second factor $x-4$ to zero} & : & x-4 &= 0 \\ \text{Solve for $x$} & : & x &= 4 \\ \text{Set the third factor $x+4$ to zero} & : & x+4 &= 0 \\ \text{Solve for $x$} & : & x &= -4 \\ \end{align*}[/latex]
The solutions are [latex]0[/latex] (double solution), [latex]4[/latex], and [latex]-4[/latex].
Solve the polynomial by factoring:
[latex]x^3+x^2-9x-9 = 0[/latex]
[latex]\begin{align*} \text{Original equation} & : & x^3 + x^2 - 9x - 9 &= 0 \\ \text{Group terms} & : & (x^3 + x^2) - (9x + 9) &= 0 \\ \text{Factor out the common factors in each group} & : & x^2(x + 1) - 9(x + 1) &= 0 \\ \text{Factor by grouping} & : & (x^2 - 9)(x + 1) &= 0 \\ \text{Factor $x^2 - 9$ as a difference of squares} & : & (x - 3)(x + 3)(x + 1) &= 0 \\ \end{align*}[/latex]
Applying zero-product property:
[latex]\begin{align*} \text{Set first factor $x-3$ to zero} & : & x - 3 &= 0 & \Rightarrow x &= 3 \\ \text{Set second factor $x+3$ to zero} & : & x + 3 &= 0 & \Rightarrow x &= -3 \\ \text{Set third factor $x+1$ to zero} & : & x + 1 &= 0 & \Rightarrow x &= -1 \\ \end{align*}[/latex]
The solutions are [latex]3, -3, \text{ and }-1[/latex].