Algebra II: Quadratic Equations - Factoring (Level 1 of 10)

Algebra II: Quadratic Equations - Factoring (Level 1 of 10)

Assessment

Interactive Video

Mathematics

11th Grade - University

Hard

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The video tutorial covers the basics of factoring quadratic equations, emphasizing the zero product property. It explains that not all quadratic equations can be factored over integers and highlights the importance of other techniques for finding solutions. The tutorial provides multiple examples of solving quadratic equations by factoring, demonstrating the process step-by-step and addressing common misconceptions.

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7 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does it mean to factor a polynomial?

To divide a polynomial by a constant

To subtract one polynomial from another

To write a polynomial as a product of other polynomials

To add polynomials together

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which of the following is true about quadratic equations that cannot be factored over integers?

They have no solutions

They can only have real solutions

They can be solved by factoring over integers

They may have real, imaginary, or complex solutions

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the zero product property used for?

To find the sum of two numbers

To solve equations by setting each factor to zero

To multiply two polynomials

To divide a polynomial by zero

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

When solving the equation (y + 4)(y + 6) = 0, what are the solutions for y?

y = 2 and y = 3

y = 0 and y = 10

y = -4 and y = -6

y = 4 and y = 6

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is a repeated root in the context of quadratic equations?

A root that appears twice in the solution

A root that cannot be factored

A root that is imaginary

A root that is complex

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the equation x(x - 7) = 0, what are the solutions for x?

x = 7 and x = 14

x = 0 and x = -7

x = 0 and x = 7

x = 7 and x = -7

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

When solving 3z(2z + 5) = 0, what is one of the solutions for z?

z = 3

z = 0

z = 5/2

z = -5/2