Common Core: High School - Algebra : Rewriting Expressions Using its Structure: CCSS.Math.Content.HSA-SSE.A.2

Study concepts, example questions & explanations for Common Core: High School - Algebra

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All Common Core: High School - Algebra Resources

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Example Questions

Example Question #11 : Rewriting Expressions Using Its Structure: Ccss.Math.Content.Hsa Sse.A.2

Factor the quadratic expression.

\(\displaystyle x^2-9\)

Possible Answers:

\(\displaystyle (x-3)(x-3)\)

\(\displaystyle (x-3)(x+3)\)

\(\displaystyle (x-9)(x+1)\)

\(\displaystyle (x+3)(x+3)\)

\(\displaystyle (x-1)(x+9)\)

Correct answer:

\(\displaystyle (x-3)(x+3)\)

Explanation:

To factor the quadratic expression 

\(\displaystyle x^2-9\)

set up the expression in factored form, leaving blanks for the numbers that are not yet known.

\(\displaystyle (x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})\)

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-9, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-9) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 3 and -3, as the product of 3 and -3 is -9, and sum of 3 and -3 is 0. So, this results in the expression's factored form looking like...

\(\displaystyle (x-3)(x+3)\)

This is known as a difference of squares.

Example Question #32 : Seeing Structure In Expressions

Factor the quadratic expression.

\(\displaystyle x^2-16\)

Possible Answers:

\(\displaystyle (x-4)(x+4)\)

\(\displaystyle (x-8)(x+2)\)

\(\displaystyle (x-2)(x+8)\)

\(\displaystyle (x+4)(x+4)\)

\(\displaystyle (x-4)(x-4)\)

Correct answer:

\(\displaystyle (x-4)(x+4)\)

Explanation:

To factor the quadratic expression 

\(\displaystyle x^2-16\)

set up the expression in factored form, leaving blanks for the numbers that are not yet known.

\(\displaystyle (x+\underline{\hspace{.5cm}})(x+\underline{\hspace{.5cm}})\)

At this point, you need to find two numbers - one for each blank. By looking at the original expression, a few clues can be gathered that will help find the two numbers. The product of these two numbers will be equal to the last term of the original expression (-16, or c in the standard quadratic formula), and their sum will be equal to the coefficient of the second term of the original expression (0, or b in the standard quadratic formula). Because their product is negative (-16) and the sum is zero, that must mean that they have different signs but the same absolute value.

Now, at this point, test a few different possibilities using the clues gathered from the original expression. In the end, it's found that the only numbers that work are 4 and -4, as the product of 4 and -4 is -16, and sum of 4 and -4 is 0. So, this results in the expression's factored form looking like...

\(\displaystyle (x-4)(x+4)\)

This is known as a difference of squares.

All Common Core: High School - Algebra Resources

8 Diagnostic Tests 97 Practice Tests Question of the Day Flashcards Learn by Concept
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