Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

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 $9 x^{2} + 21 x + 10$ .

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

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

$= (3 x)^{2} + 7 (3 x) + 10$

Step 3 - Substitute $N$ for the $3 x$ .

$= N^{2} + 7 N + 10$

Step 4 - Factor the new trinomial.

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

Step 5 - Replace $N$ with $3 x$ to find the final factored form.

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

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 $(3 x + 2) (3 x + 5)$ . Write the resulting expression in standard form. Be prepared to explain if it is equivalent to $9 x^{2} + 21 x + 10$ .

Enter the simplified expression and your explanation.

Compare your answer:

$(3 x + 2) (3 x + 5) = 9 x^{2} + 15 x + 6 x + 10 = 9 x^{2} + 21 x + 10$ . Yes, the expressions are equivalent.

2.

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

Compare your answer:

The trinomial using $N$ has a squared term with a coefficient of 1. Its factors can be found more easily.

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

3.

$4 x^{2} + 28 x + 45$

Enter the expression in factored form.

Compare your answer:

The square root of the first term, $2 x$ , is a factor of the second term, $28 x$ . The expression can be rewritten as $(2 x)^{2} + 14 (2 x) + 45$ . Substitute $N$ for the $2 x$ : $N^{2} + 14 N + 45$ . This new trinomial can be factored into $(N + 5) (N + 9)$ . Replace $N$ with $2 x$ to find the final factored form: $(2 x + 5) (2 x + 9)$ . So, $4 x^{2} + 28 x + 45 = (2 x + 5) (2 x + 9)$ .

4.

$25 x^{2} - 35 x + 6$

Enter the expression in factored form.

Compare your answer:

The square root of the first term, $5 x$ , is a factor of the second term, $35 x$ . The expression can be rewritten as $(5 x)^{2} - 7 (5 x) + 6$ . Substitute $N$ for the $5 x$ : $N^{2} - 7 N + 6$ . This new trinomial can be factored into $(N - 6) (N - 1)$ . Replace $N$ with $5 x$ to find the final factored form: $(5 x - 6) (5 x - 1)$ . So, $25 x^{2} - 35 x + 6 = (5 x - 6) (5 x - 1)$ .

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.

$5 x^{2} + 17 x + 6$

We see that multiplying the leading coefficient by 5 would make the first term into a perfect square, $25 x^{2}$ . 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 $\frac{1}{5}$ , since $\frac{1}{5} \cdot 5$ is the same as multiplying by 1.

Step 1 - Multiply by $\frac{1}{5} \cdot 5$ .

$= \frac{1}{5} \cdot 5 (5 x^{2} + 17 x + 6)$

Step 2 - Distribute the 5 into the trinomial.

$= \frac{1}{5} (25 x^{2} + 85 x + 30)$

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

$= \frac{1}{5} ((5 x)^{2} + 17 (5 x) + 30)$

Step 4 - Substitute $N$ for $5 x$ .

$= \frac{1}{5} (N^{2} + 17 N + 30)$

Step 5 - Factor the trinomial.

$= \frac{1}{5} (N + 15) (N + 2)$

Step 6 - Replace $N$ with $5 x$ to find the final factored form.

$= \frac{1}{5} (5 x + 15) (5 x + 2)$

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

$= \frac{1}{5} \cdot 5 (x + 3) (5 x + 2)$

Step 8 - $\frac{1}{5} \cdot 5 = 1$

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

5.

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

Enter the expression in standard form.

Compare your answer:

$(x + 3) (5 x + 2) = 5 x^{2} + 2 x + 15 x + 6 = 5 x^{2} + 17 x + 6$ . Yes, the expressions are equivalent.

6.

Applying the same strategy from the example to a trinomial with a leading term of $6 x^{2}$ , what is the multiplication expression equivalent to 1 you would use to rewrite the trinomial? Be prepared to show your reasoning.

Compare your answer:

$\frac{1}{6} \cdot 6$ ; multiplying the leading term of the trinomial by 6 would result in a leading term of $36 x^{2}$ , which is a perfect square.

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

7.

$3 x^{2} + 16 x + 5$

Enter the expression in factored form.

Compare your answer:

Multiply the expression by $\frac{1}{3} \cdot 3$ .

Distributing the 3 into the trinomial gives $\frac{1}{3} (9 x^{2} + 48 x + 15)$ .

Rewriting the expression in terms of the square root of the first term, $3 x$ , gives $\frac{1}{3} ((3 x)^{2} + 16 (3 x) + 15)$ .

Substituting $N$ for $3 x$ makes the expression $\frac{1}{3} (N^{2} + 16 N + 15)$ , which can be factored into $\frac{1}{3} (N + 15) (N + 1)$ .

$N$ can now be replaced with $3 x$ and the expression becomes $\frac{1}{3} (3 x + 15) (3 x + 1)$ .

Factor 3 out of the first factor, because $\frac{1}{3} \cdot 3 = 1$ . $\frac{1}{3} (3 x + 15) (3 x + 1) = (x + 5) (3 x + 1)$ So, $3 x^{2} + 16 x + 5 = (x + 5) (3 x + 1)$ .

8.

$10 x^{2} - 41 x + 4$

Enter the expression in factored form.

Compare your answer:

Multiply the expression by $\frac{1}{10} \cdot 10$ .

Distributing the 10 into the trinomial gives $\frac{1}{10} (100 x^{2} - 410 x + 40)$ . Rewriting the expression in terms of the square root of the first term, $10 x$ , gives $\frac{1}{10} ((10 x)^{2} - 41 (10 x) + 40)$ . Substituting $N$ for $10 x$ makes the expression $\frac{1}{10} (N^{2} - 41 N + 40)$ , which can be factored into $\frac{1}{10} (N - 40) (N - 1)$ . $N$ can now be replaced with $10 x$ and the expression becomes $\frac{1}{10} (10 x - 40) (10 x - 1)$ . Factor 10 out of the first factor, because $\frac{1}{10} \cdot 10 = 1$ . $\frac{1}{10} (10 x - 40) (10 x - 1) = (x - 4) (10 x - 1)$ So, $10 x^{2} - 41 x + 4 = (x - 4) (10 x - 1)$ .

Self Check

Find the factored form of the quadratic expression

9x^2− 6x − 35

.

$(3x + 7)(3x - 5)$
$(x + 7)(x - 5)$
$(3x - 7)(3x + 5)$
$(9x + 7)(x - 5)$

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

$4 x^{2} + 4 x - 3$

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

$= (2 x)^{2} + 2 (2 x) - 3$

Step 2 - Substitute $N$ for $2 x$ .

$= N^{2} + 2 N - 3$

Step 3 - Factor the simplified expression.

$= (N + 3) (N - 1)$

Step 4 - Replace $N$ with $2 x$ .

$= (2 x + 3) (2 x - 1)$

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

$5 x^{2} + 27 x - 18$

Step 1 - Multiply by $\frac{1}{5} \cdot 5$ to make a coefficient with a squared term without impacting the overall value of the expression.

$= \frac{1}{5} \cdot 5 (5 x^{2} + 27 x - 18)$

Step 2 - Distribute the 5.

$= \frac{1}{5} (25 x^{2} + 135 x - 90)$

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

$= \frac{1}{5} ((5 x)^{2} + 27 (5 x) - 90)$

Step 4 - Substitute $N$ for $5 x$ .

$= \frac{1}{5} (N^{2} + 27 N - 90)$

Step 5 - Factor the simplified expression.

$= \frac{1}{5} (N + 30) (N - 3)$

Step 6 - Replace $N$ with $5 x$ .

$= \frac{1}{5} (5 x + 30) (5 x - 3)$

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

$= \frac{1}{5} \cdot 5 (x + 6) (5 x - 3)$

Step 8 - $\frac{1}{5} \cdot 5 = 1$

$= (x + 6) (5 x - 3)$

Step 9 - Use the distributive property, or FOIL, to expand $(x + 6) (5 x - 3)$ . Is it equivalent to $5 x^{2} + 27 x - 18$ ?

$(x + 6) (5 x - 3)$

$= 5 x^{2} + 30 x - 3 x - 18$

$= 5 x^{2} + 27 x - 18$

They are equivalent. $✓$

Try it

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

Find the factored form of $3 x^{2} + 25 x + 42$ .

Here is how to use the shortcuts you just learned to find the factored form of a quadratic equation:

$3 x^{2} + 25 x + 42$

Step 1 - Multiply by $\frac{1}{3} \cdot 3$ to make a coefficient with a squared term without impacting the overall value of the expression. $= \frac{1}{3} \cdot 3 (3 x^{2} + 25 x + 42)$

Step 2 - Distribute the 3.

$= \frac{1}{3} (9 x^{2} + 75 x + 126)$

Step 3 - Rewrite the first two terms with $3 x$ as the common factor.

$= \frac{1}{3} ((3 x)^{2} + 25 (3 x) + 126)$

Step 4 - Substitute $N$ for $3 x$ .

$= \frac{1}{3} (N^{2} + 25 N + 126)$

Step 5 - Factor the simplified expression.

$= \frac{1}{3} (N + 18) (N + 7)$

Step 6 - Replace $N$ with $3 x$ .

$= \frac{1}{3} (3 x + 18) (3 x + 7)$

Step 7 - Factor 3 out of one of the expressions.

$= \frac{1}{3} (3 x + 18) (3 x + 7)$

Step 8 - $\frac{1}{3} \cdot 3 = 1$

$= (x + 6) (3 x + 7)$

The factored form is $(x + 6) (3 x + 7)$ .

8.10.4 Finding the Factors of Quadratic Expressions in Standard Form

Activity

Additional Resources

Finding the Factors of Quadratic Expressions in Standard Form

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