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Algebra 1

8.10.4 Finding the Factors of Quadratic Expressions in Standard Form

Algebra 18.10.4 Finding the Factors of Quadratic Expressions in Standard Form

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Activity

This activity is an extension of the expectations in the TEKS.

Here is a clever way to think about quadratic expressions that can make it easier to find their factored form.

Factor 9x2+21x+109x2+21x+10.

Step 1 - Notice the square root of the first term, 3x3x, is a factor of the second term.

Step 2 - Rewrite in terms of 3x3x. The third term is unchanged.

=(3x)2+7(3x)+10=(3x)2+7(3x)+10

Step 3 - Substitute NN for the 3x3x.

=N2+7N+10=N2+7N+10

Step 4 - Factor the new trinomial.

=(N+2)(N+5)=(N+2)(N+5)

Step 5 - Replace NN with 3x3x to find the final factored form.

=(3x+2)(3x+5)=(3x+2)(3x+5)

Using this substitution method creates a simpler trinomial that we may factor more easily.

1.

Use the distributive property, or FOIL, to expand and simplify (3x+2)(3x+5)(3x+2)(3x+5). Write the resulting expression in standard form. Be prepared to explain if it is equivalent to 9x2+21x+109x2+21x+10.

2.

Study the substitution method and make sense of what was done in each step. Why might the simplified trinomial using NN be simpler to factor?

In questions 3 and 4, try this NN-substitution method to write each of these expressions in factored form.

3.

4 x 2 + 28 x + 45 4 x 2 + 28 x + 45

4.

25 x 2 35 x + 6 25 x 2 35 x + 6

You have probably noticed that the coefficient of the squared term in all of the previous examples is a perfect square. What if that coefficient is not a perfect square?

Here is an example of an expression whose squared term has a coefficient that is not a squared term.

5x2+17x+65x2+17x+6

We see that multiplying the leading coefficient by 5 would make the first term into a perfect square, 25x225x2. But, we cannot introduce a new factor that will impact the value of the expression.

We need to find a way to balance the impact of multiplying everything by 5.

Knowing this, we multiply the entire trinomial by 1515 , since 15·515·5 is the same as multiplying by 1.

Step 1 - Multiply by 15·515·5.

=15·5(5x2+17x+6)=15·5(5x2+17x+6)

Step 2 - Distribute the 5 into the trinomial.

=15(25x2+85x+30)=15(25x2+85x+30)

Step 3 - Rewrite in terms of the square root of the first term, 5x5x.

=15((5x)2+17(5x)+30)=15((5x)2+17(5x)+30)

Step 4 - Substitute NN for 5x5x.

=15(N2+17N+30)=15(N2+17N+30)

Step 5 - Factor the trinomial.

=15(N+15)(N+2)=15(N+15)(N+2)

Step 6 - Replace NN with 5x5x to find the final factored form.

=15(5x+15)(5x+2)=15(5x+15)(5x+2)

Step 7 - Factor 5 out of one of the factors.

=15·5(x+3)(5x+2)=15·5(x+3)(5x+2)

Step 8 - 15·5=115·5=1

=(x+3)(5x+2)=(x+3)(5x+2)

5.

Use the distributive property, or FOIL, to expand and simplify (x+3)(5x+2)(x+3)(5x+2). Write the resulting expression in standard form. Be prepared to explain if it is equivalent to 5x2+17x+65x2+17x+6.

6.

Applying the same strategy from the example to a trinomial with a leading term of 6x26x2, what is the multiplication expression equivalent to 1 you would use to rewrite the trinomial? Be prepared to show your reasoning.

For questions 7 and 8, try the NN-method to write each of these expressions in factored form.

7.

3 x 2 + 16 x + 5 3 x 2 + 16 x + 5

8.

10 x 2 41 x + 4 10 x 2 41 x + 4

Self Check

Find the factored form of the quadratic expression 9 x 2 6 x 35 .
  1. ( 3 x + 7 ) ( 3 x 5 )
  2. ( x + 7 ) ( x 5 )
  3. ( 3 x 7 ) ( 3 x + 5 )
  4. ( 9 x + 7 ) ( x 5 )

Additional Resources

Finding the Factors of Quadratic Expressions in Standard Form

Let's look at a shortcut that can help to factor a complicated quadratic expression. This method will be useful only when the coefficient of the first term is a perfect square, such as 4 or 9.

Example 1

4x2+4x34x2+4x3

Step 1 - Rewrite the first two terms with 2x2x as the common factor.

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

Step 2 - Substitute NN for 2x2x.

=N2+2N3=N2+2N3

Step 3 - Factor the simplified expression.

=(N+3)(N1)=(N+3)(N1)

Step 4 - Replace NN with 2x2x.

=(2x+3)(2x1)=(2x+3)(2x1)

Here is an example of an expression whose squared term has a coefficient that is not a squared term. A different method can still be applied.

Example 2

5x2+27x185x2+27x18

Step 1 - Multiply by 15·515·5 to make a coefficient with a squared term without impacting the overall value of the expression.

=15·5(5x2+27x18)=15·5(5x2+27x18)

Step 2 - Distribute the 5.

=15(25x2+135x90)=15(25x2+135x90)

Step 3 - Rewrite the first two terms with 5x5x as the common factor.

=15((5x)2+27(5x)90)=15((5x)2+27(5x)90)

Step 4 - Substitute NN for 5x5x.

=15(N2+27N90)=15(N2+27N90)

Step 5 - Factor the simplified expression.

=15(N+30)(N3)=15(N+30)(N3)

Step 6 - Replace NN with 5x5x.

=15(5x+30)(5x3)=15(5x+30)(5x3)

Step 7 - Factor 5 out of one of the expressions.

=15·5(x+6)(5x3)=15·5(x+6)(5x3)

Step 8 - 15·5=115·5=1

=(x+6)(5x3)=(x+6)(5x3)

Step 9 - Use the distributive property, or FOIL, to expand (x+6)(5x3)(x+6)(5x3). Is it equivalent to 5x2+27x185x2+27x18?

(x+6)(5x3)(x+6)(5x3)

=5x2+30x3x18=5x2+30x3x18

=5x2+27x18=5x2+27x18

They are equivalent.

Try it

Try It: Finding the Factors of Quadratic Expressions in Standard Form

Find the factored form of 3x2+25x+423x2+25x+42.

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