Zero-Door DoormanTilewrights · polynomials and factoring

Two doors, and one of them must be open: the zero product property.

Solving Quadratics by Factoring

The zero product property. In MathBound, the free Algebra I adventure game, Zero-Door Doorman asks these problems, and the only way past is to solve them.

Part of Polynomials and Factoring: chapter 8 of the standard course.

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10 questions, no sign-up, nothing saved. Climbs from the basics to harder questions.

What you will practice

Zero Product Property. Solve an equation that is already factored and set to zero.
Solve by Factoring. Solve a quadratic equation already in standard form (set to zero) with a leading coefficient of 1.
Solve Non-Standard Equations. Solve a quadratic equation requiring terms to be moved to one side to set it to zero.
Solve Higher Degree Equations. Solve a cubic equation by first factoring out a variable GCF to reveal a quadratic.
Applications. Solve a geometry application where negative solutions must be discarded (extraneous roots).

Worked examples

Made by the game’s own question engine. Try each one, then open its steps.

Example 1 · Zero Product Property

Which

equation

has

the

solutions

x = - 8

and

x = - 1 ?

$(x - 8) (x + 1) = 0$
$(x - 8) (x - 1) = 0$
$(x + 8) (x + 1) = 0$
$(x + 8) (x - 1) = 0$

Show the worked steps

$x = - 8$ $comes$ $from$ $the$ $factor$ $(x + 8),$ $and$ $x = - 1$ $from$ $(x + 1) .$
$So$ $the$ $equation$ $is$ $(x + 8) (x + 1) = 0 .$

Answer: $(x + 8) (x + 1) = 0$

Example 2 · Solve by Factoring

Solve:

x^{2} + 2 x - 35 = 0

$x = - 2$
$x = - 35$
$x = - 7$ $or$ $x = 5$
$x = - 5$ $or$ $x = 7$

Show the worked steps

$Factor:$ $(x + 7) (x - 5) = 0.$
$Each$ $factor$ $gives$ $a$ $root:$ $x = - 7$ $and$ $x = 5 .$

Answer: $x = - 7$ $or$ $x = 5$

Example 3 · Solve Non-Standard Equations

Lena

solved

(x + 2) (x - 3) = - 7 (x + 2)

by

dividing

both

sides

by

x + 2

and

got

x = - 4 .

Which

solution

was

lost,

and

why?

$x = 4 :$ $a$ $square$ $root$ $has$ $two$ $signs$
$x = - 2 :$ $there$ $x + 2 = 0,$ $and$ $dividing$ $by$ $zero$ $is$ $not$ $allowed$
$x = 2 :$ $there$ $x + 2 = 0,$ $and$ $dividing$ $by$ $zero$ $is$ $not$ $allowed$
$No$ $solution$ $was$ $lost:$ $dividing$ $both$ $sides$ $by$ $the$ $same$ $thing$ $keeps$ $them$ $equal$

Show the worked steps

$Dividing$ $by$ $x + 2$ $is$ $only$ $allowed$ $when$ $x + 2$ $is$ $not$ $zero,$ $so$ $x = - 2$ $has$ $to$ $be$ $checked$ $on$ $its$ $own:$ $it$ $works.$
$Safer:$ $move$ $everything$ $to$ $one$ $side$ $and$ $factor:$ $(x + 2) (x + 4) = 0,$ $so$ $x = - 2$ $or$ $x = - 4 .$

Answer: $x = - 2 :$ $there$ $x + 2 = 0,$ $and$ $dividing$ $by$ $zero$ $is$ $not$ $allowed$

Where it is in your textbook

Textbook	Lessons
Standard course	8.8, 11.3
Big Ideas Math Algebra 1	7.4, 7.5, 7.6, 7.8, 8.5, 10.3
Big Ideas Math Algebra 1 (2022)	7.4, 7.5, 7.6, 7.8, 8.5, 10.3
Carnegie Learning Algebra I	M5 2.2, M5 2.4, M5 3.2
CPM Core Connections Algebra	8.2.2, 9.1.1, 10.2.1, 10.3.2
CPS Skyline Algebra	7.12-7.13
enVision Algebra 1	9-2
enVision Algebra 1 (2024)	9-2
Eureka Math Algebra I (EngageNY)	1.17-1.19, 4.5, 4.6-4.7
Glencoe Algebra 1	8-5, 8-6, 8-7, 10-4
HMH Into Algebra 1	17.2, 17.3, 17.4, 20.3, 20.4
Holt McDougal Larson Algebra 1	8.4, 8.5, 8.6, 8.8
Illustrative Mathematics Algebra 1	7.4-7.5, 7.9-7.10
Illustrative Mathematics Algebra 1 (v.360)	8.4-8.5, 8.9-8.10
Math Nation Algebra 1 (Florida B.E.S.T.)	9.1, 9.7, 9.10, 9.11
Open Up NC Math 1 (MVP)	7.6, 7.8
OpenStax Elementary Algebra 2e	7.6, 9.6
Pearson Algebra 1 Common Core	9-4, 10-4
Reveal Algebra 1	11-4
SpringBoard Algebra 1	31, 35

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