Search topics

Search

Solve a higher-degree polynomial equation or inequality

Start a set of 8
Step 5 of 5

Using the null factor law too early. One of these lines is right. Which one?

Year 11, VCE Units 1 & 2, Chapter 5, 4 difficulty levels, No calculatorBuild a worksheet

Higher-degree equations and inequalities

Put the polynomial relation into zero form and factor it completely before finding roots. For an inequality, use the ordered roots to test signs on intervals and apply the strict or inclusive endpoint rule

\[A(x)B(x)=0\Rightarrow A(x)=0\text{ or }B(x)=0\]
  1. Compare with zero Move all terms to one side before applying the null factor law \(\displaystyle F(x)=G(x)\Rightarrow F(x)-G(x)=0\)
  2. Factor completely Keep the scale and every factor so that all real roots can be found \(\displaystyle 2\left(x-1\right)^2\left(x+3\right)=0\)
  3. Apply the null factor law If a product is zero, at least one of its factors is zero \(\displaystyle A(x)B(x)=0\Rightarrow A(x)=0\text{ or }B(x)=0\)
  4. Use signs and endpoints For an inequality, test each interval and include a root only when equality is allowed \(\displaystyle F(x)\ge0\Rightarrow\text{include all roots and intervals where }F(x)>0\)

Worked example

Solve the polynomial inequality below over the real numbers

\[-2\left(x+1\right)^2\left(x-3\right)\left(x-4\right)\ge0\]
  1. \(\displaystyle x=-1\text{ with multiplicity }2,\qquad x=3\text{ with multiplicity }1,\qquad x=4\text{ with multiplicity }1\)Set each factor equal to zero and record the exponent at each root
  2. \(\displaystyle F(x)=-2\left(x+1\right)^2\left(x-3\right)\left(x-4\right),\qquad F(5)=-2(6)^2(2)(1)=-144<0\)Test a value to establish the sign on x>4
  3. \(\displaystyle F(x)<0\text{ on }x<-1\text{ and }-1<x<3;\quad F(x)>0\text{ on }3<x<4;\quad F(x)<0\text{ on }x>4\)The even root at −1 does not change the sign, while each simple root changes it
  4. \(\displaystyle F(-1)=F(3)=F(4)=0\)The factorised expression is zero at every critical root
  5. \(\displaystyle F(x)\ge0\Rightarrow x\in\left\{-1\right\}\cup\left[3,4\right]\)Include the isolated zero at −1 and the interval from 3 to 4 because the inequality is inclusive

Common mistakes

Using the null factor law too early
✕\(\displaystyle \left(x-2\right)\left(x+1\right)=3\Rightarrow x=2\text{ or }x=-1\)✓\(\displaystyle \left(x-2\right)\left(x+1\right)-3=0\Rightarrow\text{now factorise or solve the zero-form equation}\)
The null factor law requires the complete product to be equal to zero first
Changing sign at an even root
✕\(\displaystyle F(x)=\left(x+1\right)^2\left(x-3\right)\Rightarrow\text{the sign changes at }x=-1\)✓\(\displaystyle F(x)=\left(x+1\right)^2\left(x-3\right)\Rightarrow\text{the sign stays the same at }x=-1\)
An even-multiplicity root touches the axis without changing the sign of the polynomial
Excluding roots from an inclusive inequality
✕\(\displaystyle x^2-9\le0\Rightarrow-3<x<3\)✓\(\displaystyle x^2-9\le0\Rightarrow-3\le x\le3\)
A non-strict inequality allows equality, so its roots are included in the solution set

Difficulty levels

introAn already factorised equation with integer roots is solved using the null factor law
standardAn expanded quartic or a quadratic-in-a-power equation is solved exactly
challengeA repeated-root inequality or factor-theorem division stage is included
examRearrangement, factor choice, root data, signs and strict or inclusive intervals are combined

Questions about this topic

Why must an equation be written in zero form before applying the null factor law?+

The null factor law applies to a product equal to zero. Moving all terms to one side creates that required form

How does an even-multiplicity root affect an inequality sign chart?+

The sign stays the same on both sides of an even-multiplicity root because the graph touches the axis there instead of crossing it

When are roots included in an inequality solution?+

Roots are included for a non-strict relation such as ≤ or ≥, and excluded for a strict relation such as < or >

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 5: Higher-degree polynomials

In this chapter

Interpret and determine a transformed power functionConnect a factorised higher-degree rule and its graphSolve a higher-degree polynomial equation or inequalityAnalyse a parameterised polynomial familyBracket and approximate a polynomial root by bisectionApproximate solutions from intersections of two graphsLocate a polynomial turning point numerically