Math 2–3 Honors · Lesson Review · October 27, 2026

Roots, Solutions, and Zeros

Compare a polynomial expression, an equation, and a function built from the same quadratic. Use factored form to find the relevant inputs and distinguish zeros from ordered-pair solutions.

6-minute edited lesson8 chaptersCaptions + transcript

01 · Watch

Lesson video

VTT captions SRT captions

0:0001Three objects: expression, equation, function 1:2202Factor once, interpret three ways 2:2403Zeros are inputs of a function 2:3604Solutions of equations and roots of polynomials 3:4305Inputs versus ordered pairs 4:3206A nonzero scale does not change the zeros 5:0507Read the roots of a polynomial 5:2008Read the solutions of an equation

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02 · Understand

Learning targets

Identify whether the displayed object is an expression, an equation, or a function.
Find roots, equation solutions, and function zeros from factored form.
Explain why a nonzero constant factor leaves the zero inputs unchanged.
Distinguish a zero input a from the graph point (a,0).
Use the course vocabulary while recognizing that textbooks often use root and zero interchangeably.

03 · Connect

Key ideas

The lesson uses three classroom labels to keep the object and the requested answer clear. These labels help with interpretation; mathematical texts often use roots and zeros as overlapping terms.

01An expression has a value after substitution: $p(x)=x^2+2x-15,\qquad p(-5)=p(3)=0$ The roots of this polynomial are inputs that make its value zero. The expression alone is not an equation to solve, although finding its roots means solving the associated equation p(x)=0.
02An equation asks what makes the statement true: $x^2+2x-15=0\quad\Longleftrightarrow\quad x=-5\text{ or }x=3$ Each solution is a number because this equation has only the variable x. The equation must be true after that number is substituted.
03A function zero is an input with output zero: $f(a)=0\quad\Longrightarrow\quad(a,0)\text{ is on the graph}$ A zero is the input a; its x-intercept is the ordered pair (a,0). Solutions of the graph equation y=f(x) are ordered pairs, including many whose y-coordinate is not zero.
04A product is zero when a factor is zero: $c(x-r_1)(x-r_2)=0,\ c\ne0\quad\Longleftrightarrow\quad x=r_1\text{ or }x=r_2$ The nonzero constant factor changes other function values but does not create or remove zeros. Read each linear factor by setting it equal to zero; do not copy its sign.
05Root and zero are often used interchangeably: $\text{root of }p\;\Longleftrightarrow\;\text{zero of }p\;\Longleftrightarrow\;\text{solution of }p(x)=0$ In this class, expression → roots, equation → solutions, function → zeros is a useful naming convention. A textbook saying “roots of an equation” is using a common convention. Identify the mathematical condition being asked for.

04 · Apply

Worked examples

These examples follow the methods developed in the recorded lesson.

Classroom example · one quadratic, three objects

Interpret x²+2x−15 in three settings

x^2+2x-15=(x+5)(x-3)

The constants multiply to −15 and add to 2.

p(-5)=(-5+5)(-5-3)=0,\qquad p(3)=(3+5)(3-3)=0

As a polynomial expression, p has roots −5 and 3.

0=(x+5)(x-3)\quad\Longrightarrow\quad\boxed{x=-5\text{ or }x=3}

As an equation, the same numbers are solutions.

f(x)=(x+5)(x-3)\quad\Longrightarrow\quad\boxed{\text{zeros }-5,3}

As a function, the numbers are zero inputs.

\boxed{(-5,0),\ (3,0)}

These ordered pairs are the x-intercepts. For comparison, f(0)=−15 gives another graph solution (0,−15), which is not an x-intercept.

Classroom example · function

Find the zeros of f(x)=5(x+4)(x−3)

5(x+4)(x-3)=0

Set the output equal to zero. The factor 5 is nonzero.

x+4=0\quad\text{or}\quad x-3=0

Apply the zero-product property.

\boxed{\text{zeros }-4,3;\quad\text{x-intercepts }(-4,0),(3,0)}

The constant factor 5 does not change the zero inputs.

Classroom example · polynomial expression

Find the roots of (x+5)(x+2)

(x+5)(x+2)=0

Finding roots means asking when the polynomial value is zero.

x=-5\quad\text{or}\quad x=-2

Set each factor equal to zero.

\boxed{\text{roots }-5,-2}

Both constants are positive inside the factors, so both root inputs are negative.

Classroom example · equation

Solve 0=(x−2)(x+2)

x-2=0\quad\text{or}\quad x+2=0

Use the zero-product property; the equation is already set equal to zero.

\boxed{x=2\text{ or }x=-2}

Both values make the equation true. They are the roots of the associated polynomial as well.

05 · Avoid

Common mistakes

Copying the sign in a factor

The factor x+5 vanishes at x=−5, not x=5. Solve the small equation formed by setting the factor to zero.

Giving a point when an input was requested

A function zero is a number such as −5. Its x-intercept is the point (−5,0). Match the requested answer type.

Calling every graph point a zero

The graph equation y=f(x) has an ordered-pair solution for every allowed input. Only points with output zero are x-intercepts.

Using the zero-product property without zero

From (x+1)(x−5)=7, you cannot set the factors equal to zero. First bring all terms to one side and solve the resulting equation.

Treating classroom terminology as a universal restriction

Roots, zeros, and solutions overlap in many books. The condition p(a)=0 and the object being discussed matter more than a vocabulary disagreement.

06 · Check yourself

Try it first

Use these five fresh problems to check your understanding. Open a solution after you try it.

1Find the roots of this polynomial.

x^2-x-12

$x^2-x-12=(x-4)(x+3)$

The constants multiply to −12 and add to −1.

$(x-4)(x+3)=0\quad\Longrightarrow\quad\boxed{4,-3}$

These inputs make the polynomial value zero. Check: 16−4−12=0 and 9+3−12=0.

2Solve the equation.

0=2(x+1)(x-5)

$x+1=0\quad\text{or}\quad x-5=0$

The constant factor 2 is nonzero.

$\boxed{x=-1\text{ or }x=5}$

Each number makes one factor zero, so the entire right-hand side is zero.

3Find the zeros and the x-intercepts.

g(x)=-3(x-2)(x+4)

$-3(x-2)(x+4)=0\quad\Longrightarrow\quad x=2\text{ or }x=-4$

The nonzero scale −3 leaves the zero inputs unchanged.

$\boxed{\text{zeros }2,-4;\quad\text{x-intercepts }(2,0),(-4,0)}$

Separate the input numbers from the points on the graph.

4How many distinct zeros does this function have?

f(x)=x^2-4x+4

$f(x)=(x-2)^2$

Both linear factors are the same.

$\boxed{\text{one distinct zero: }2\text{, with multiplicity }2}$

The repeated factor gives the single x-intercept (2,0). It does not give two different zero inputs.

5Is (1,−6) a graph solution? Is 1 a zero?

y=x^2-x-6

$1^2-1-6=-6$

Substitution gives y=−6, so (1,−6) satisfies the graph equation.

$\boxed{(1,-6)\text{ is a graph solution; }1\text{ is not a zero.}}$

The output at 1 is −6 rather than zero.

$x^2-x-6=(x-3)(x+2)\quad\Longrightarrow\quad\text{zeros }3,-2$

The graph x-intercepts are (3,0) and (−2,0).

07 · Revisit

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