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Algebra II/Irrationals & Radicals/Solving Polynomial Equations
Algebra II Regents in 15 days
Algebra II · Lesson 4

Solving Polynomial Equations

Every zero of a polynomial is a factor waiting to be found.


A structural engineer designing a roller coaster models the height of a section of track with the equation h(x)=x3x26xh(x) = x^3 - x^2 - 6x, where xx is horizontal distance in meters. She needs to know exactly where the track is at ground level — where h(x)=0h(x) = 0. Plugging numbers in randomly would take forever. Factoring the polynomial finds every answer at once. That technique is called solving a polynomial equation by finding its zeros.

A zero of a polynomial is any value of xx that makes the polynomial equal to zero. On a graph, zeros are the xx-intercepts. The connection between zeros and factors comes from a single idea called the Zero Product Property: if a product of factors equals zero, then at least one of those factors must be zero. In symbols, if ab=0ab = 0, then a=0a = 0 or b=0b = 0.

That property is the engine behind everything in this lesson. Once you factor a polynomial completely, you set each factor equal to zero and solve. Every factor gives you one equation. Every equation gives you one zero.

The process has three stages. First, factor out any greatest common factor. Second, factor the remaining expression completely — using difference of squares, trinomial factoring, or grouping depending on the form. Third, apply the Zero Product Property to each factor.

Here is a straightforward example with a quadratic first.

Solving x² + 5x + 6 = 0
x2+5x+6=0x^2 + 5x + 6 = 0
The equation is already set equal to zero — that's exactly where you need it before factoring.
(x+2)(x+3)=0(x + 2)(x + 3) = 0
Find two numbers that multiply to 6 and add to 5. Those are 2 and 3.
x+2=0orx+3=0x + 2 = 0 \quad \text{or} \quad x + 3 = 0
Zero Product Property: if the product is zero, one of the factors must be zero.
x=2orx=3x = -2 \quad \text{or} \quad x = -3 \checkmark
Solve each mini-equation. These are the two zeros.

Now a cubic — this is where the GCF step becomes essential.

Solving x³ - x² - 6x = 0
x3x26x=0x^3 - x^2 - 6x = 0
Before anything else, look for a common factor in every term.
x(x2x6)=0x(x^2 - x - 6) = 0
Every term has an x. Factor it out. Now you have a product of two factors.
x(x3)(x+2)=0x(x - 3)(x + 2) = 0
Factor the trinomial: find two numbers that multiply to -6 and add to -1. Those are -3 and 2.
x=0orx3=0orx+2=0x = 0 \quad \text{or} \quad x - 3 = 0 \quad \text{or} \quad x + 2 = 0
Three factors means three equations. Apply the Zero Product Property to all three.
x=0orx=3orx=2x = 0 \quad \text{or} \quad x = 3 \quad \text{or} \quad x = -2 \checkmark
Three zeros. The track from the engineering problem is at ground level at x = 0, x = 3, and x = -2.

The graph of a polynomial makes the zeros visible. Each zero is a point where the curve crosses or touches the xx-axis.

Interactive graph — scroll to zoom, drag to pan

The three xx-intercepts at x=2x = -2, x=0x = 0, and x=3x = 3 match the zeros found by factoring. The graph confirms the algebra.

Some polynomial equations require rearranging before you can factor. The equation must equal zero on one side before the Zero Product Property applies. If you have x2+5x=6x^2 + 5x = -6, move everything to one side first.

Solving x² + 5x = -6
x2+5x+6=0x^2 + 5x + 6 = 0
Add 6 to both sides. You need zero on one side before factoring.
(x+2)(x+3)=0(x + 2)(x + 3) = 0
Same factoring as before.
x=2orx=3x = -2 \quad \text{or} \quad x = -3 \checkmark
Same zeros. The rearrangement did not change the answers — it just put the equation in a form you can work with.

Higher-degree polynomials can also factor using a difference of squares pattern. Recall that a2b2=(a+b)(ab)a^2 - b^2 = (a+b)(a-b). This applies to polynomial terms too.

Solving x⁴ - 16 = 0
x416=0x^4 - 16 = 0
This is a difference of squares: x⁴ is (x²)² and 16 is 4².
(x2+4)(x24)=0(x^2 + 4)(x^2 - 4) = 0
Apply the difference of squares pattern.
(x2+4)(x+2)(x2)=0(x^2 + 4)(x + 2)(x - 2) = 0
Factor x² - 4 again — it's another difference of squares. The factor x² + 4 cannot be factored further over the real numbers.
x2+4=0orx+2=0orx2=0x^2 + 4 = 0 \quad \text{or} \quad x + 2 = 0 \quad \text{or} \quad x - 2 = 0
Set each factor equal to zero.
x=2orx=2x = 2 \quad \text{or} \quad x = -2 \checkmark
x² + 4 = 0 gives x² = -4, which has no real solutions. So there are only two real zeros.
Practice Questions
x27x+12=0x^2 - 7x + 12 = 0
2x3+6x28x=02x^3 + 6x^2 - 8x = 0
x3+3x2=4xx^3 + 3x^2 = 4x
Regents Corner

On Parts II and III of the Algebra II Regents, polynomial equation problems often require you to show all work. Setting each factor equal to zero must appear explicitly in your solution. Writing only the final answers without the Zero Product Property step will cost you method points even if the answers are correct. Always write out every equation before solving it.

Students move a term to the wrong side and write something like x² + 5x = -6, then factor as x(x + 5) = -6 and conclude that x = -6 or x + 5 = -6. The Zero Product Property only works when the product equals zero — not any other number. A product of two factors equaling -6 tells you nothing useful about the individual factors. Always move everything to one side first so the equation equals zero.
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