Identifying Zeros of Perfect Square Quadratic Functions by Factoring

Identifying Zeros of Perfect Square Quadratic Functions by Factoring

Assessment

Interactive Video

Mathematics

1st - 6th Grade

Hard

Created by

Quizizz Content

FREE Resource

This video tutorial teaches how to identify zeros of perfect square quadratic functions by factoring. It begins with an introduction to the perfect square pattern and explains how to factor such quadratics. The tutorial then covers finding zeros by setting the factored form equal to zero and solving for x. It also discusses the graphical representation of zeros and provides a general formula for finding them. Common mistakes in identifying zeros are highlighted, with corrections provided.

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7 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the factored form of the quadratic function 4x^2 + 12x + 9?

(x + 3)(4x + 3)

(2x + 3)(2x + 3)

(4x + 9)(x + 1)

(2x + 1)(2x + 9)

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the zero of the linear function f(x) = x - 3?

x = 3

x = 0

x = 1

x = -3

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do you find the zeros of a perfect square quadratic function?

By multiplying the terms

By subtracting the constant term

By adding the coefficients

By setting the function equal to zero

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the zero of the function (2x + 3)(2x + 3) = 0?

x = -2/3

x = -3/2

x = 3/2

x = 2/3

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is a common mistake when finding zeros in standard form?

Using the same sign of the constant term

Multiplying the terms incorrectly

Taking the opposite of the constant term

Dividing by the coefficient of the variable term

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the correct zero for the function 9x^2 - 3x + 1?

x = 3

x = 1/3

x = -1/3

x = -3

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do you express the zero of a function in factored form?

By subtracting the variable term

By multiplying the coefficients

By taking the opposite of the constant term and dividing by the coefficient

By adding the constant term