Understanding Quadratic Factorization Concepts

Understanding Quadratic Factorization Concepts

Assessment

Interactive Video

Mathematics

9th - 10th Grade

Practice Problem

Hard

Created by

Thomas White

FREE Resource

The video tutorial explains how to find the zeros of a function H(x) by using conjugates and long division. It starts with an overview of the problem, then demonstrates a trick using conjugates to find a quadratic factor. The tutorial proceeds with long division to find additional factors and verifies the results. Finally, it concludes by identifying all zeros of the function.

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15 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the initial task described in the video regarding the function H(x)?

To find the derivative of H(x)

To graph H(x)

To find the zeros and write it in factored form

To integrate H(x)

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the significance of using conjugates in the context of the video?

To determine the maximum value

To simplify the function

To solve a differential equation

To find a quadratic factor

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the sum of the zeros used in the trick to find the quadratic factor?

Three

Four

Two

One

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the product of the zeros used in the trick to find the quadratic factor?

Two

One

Four

Three

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the quadratic factor derived from the sum and product of the zeros?

x^2 + 2x + 4

x^2 - 2x + 4

x^2 - 4x + 2

x^2 + 4x - 2

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the relationship between the sum and product of zeros and the quadratic factor?

They are unrelated

They determine the coefficients of the quadratic factor

They simplify the quadratic factor

They are used to verify the factorization

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why was long division necessary in the process described in the video?

Because it simplifies the function

Because it was a requirement of the problem

Because direct factoring was not possible

Because the function was already factored

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