Understanding Factoring Quadratic Expressions Using Perfect Square Trinomials and More

Perfect Square Trinomial	Square of Binomial
$a^{2} + 2 a b + b^{2}$	$(a + b)^{2}$
$a^{2} - 2 a b + b^{2}$	$(a - b)^{2}$

Perfect Square Trinomial

Square of Binomial

a^{2} + 2 a b + b^{2}

(a + b)^{2}

a^{2} - 2 a b + b^{2}

(a - b)^{2}

Expression	First Term	Last Term
$49 x^{2} - 56 x + 16$	$49 x^{2} = (7 x)^{2}$	$16 = (4)^{2}$
$16 x^{2} + 24 x + 9$	$16 x^{2} = (4 x)^{2}$	$9 = (3)^{2}$

Expression

First Term

Last Term

49 x^{2} - 56 x + 16

49 x^{2} = (7 x)^{2}

16 = (4)^{2}

16 x^{2} + 24 x + 9

16 x^{2} = (4 x)^{2}

9 = (3)^{2}

Expression	First Term	Last Term	Middle Term
$49 x^{2} - 56 x + 16$	$49 x^{2} = (7 x)^{2}$	$16 = (4)^{2}$	$56 x = 2 \cdot 7 x \cdot 4$
$16 x^{2} + 24 x + 9$	$16 x^{2} = (4 x)^{2}$	$9 = (3)^{2}$	$24 x = 2 \cdot 4 x \cdot 3$

Expression

First Term

Last Term

Middle Term

49 x^{2} - 56 x + 16

49 x^{2} = (7 x)^{2}

16 = (4)^{2}

56 x = 2 \cdot 7 x \cdot 4

16 x^{2} + 24 x + 9

16 x^{2} = (4 x)^{2}

9 = (3)^{2}

24 x = 2 \cdot 4 x \cdot 3

Square	Side Length, $s$	Area, $s^{2}$
Larger Square	$4 n + 1$	$(4 n + 1)^{2}$
Smaller Square	$5$	$5^{2}$

Square

Side Length,

s

Area,

s^{2}

Larger Square

4 n + 1

(4 n + 1)^{2}

Smaller Square

5

5^{2}

Expression	First Term	Last Term
$25 x^{2} - 30 x + 9$	$25 x^{2} = (5 x)^{2}$	$9 = (3)^{2}$
$x^{2} - 14 x + 49$	$x^{2} = (x)^{2}$	$49 = (7)^{2}$

Expression

First Term

Last Term

25 x^{2} - 30 x + 9

25 x^{2} = (5 x)^{2}

9 = (3)^{2}

x^{2} - 14 x + 49

x^{2} = (x)^{2}

49 = (7)^{2}

Expression	First Term	Last Term	Middle Term
$25 x^{2} - 30 x + 9$	$25 x^{2} = (5 x)^{2}$	$9 = (3)^{2}$	$30 x = 2 \cdot 5 x \cdot 3$
$x^{2} - 14 x + 49$	$x^{2} = (x)^{2}$	$49 = (7)^{2}$	$14 x = 2 \cdot x \cdot 7$

Expression

First Term

Last Term

Middle Term

25 x^{2} - 30 x + 9

25 x^{2} = (5 x)^{2}

9 = (3)^{2}

30 x = 2 \cdot 5 x \cdot 3

x^{2} - 14 x + 49

x^{2} = (x)^{2}

49 = (7)^{2}

14 x = 2 \cdot x \cdot 7

Factor the following polynomial.

x^{2} - 81

25 y^{2} - 1

7 - 0.25 z^{2}

To factor the given expression, we will recall the formula for the difference of two squares. a^2- b^2=( a+ b)( a- b) Now, we will rewrite our terms as a second power. Then we will apply the formula.

Notice that the given expression can be written as the product of two conjugate binomials, (x+9) and (x-9).

This time we will rewrite 25y^2-1 in the form a^2- b^2. Let's first write 25y^2 as ( 5x)^2 and then we need also to rewrite 1 as 1^2.

The given expression is equal to the product of conjugate binomials (5y+1) and (5y-1).

Finally, we will factor the expression 7-0.25z^2. We can do this by finding the number that is equal to 7 when it is multiplied by itself. sqrt(7) * sqrt(7) = 7 Having the fact that the second power of sqrt(7) is equal to 7, we can rewrite our expression.

The expression is equal to the product of conjugate binomials (sqrt(7)+0.5z) and (sqrt(7)-0.5z).

The given expressions represent the factored form of the difference of perfect squares or perfect square trinomials.

16 y^{2} - △ = (♢ + 7) (4 y - ♡)

Which option is correct for all the missing values in place of the symbols?

A . B . C . D . △ = 7, ♢ = 16 y, ♡ = - 7 △ = 49, ♢ = 4 y, ♡ = 7 △ = 49, ♢ = 4, ♡ = - 7 △ = 7, ♢ = 16, ♡ = 7

9 x^{2} + △ + 25 = (♢ + ♡)^{2}

Which option is correct for all the missing values in place of the symbols?

A . B . C . D . △ = 90 x, ♢ = 9 x, ♡ = 5 △ = 225 x, ♢ = 15, ♡ = 625 △ = 30 x, ♢ = 3 x, ♡ = 5 △ = 150, ♢ = 6 x, ♡ = 25

Let's recall that the formula for the difference of two squares is a^2- b^2=( a+ b)( a- b). The given equation is written in that same form. 16y^2-△=( ◊ + 7)( 4y- ♡) With this in mind, if we match the first terms of the binomials, we can find that ◊= 4y. Then, if we match the second terms of the binomials, we can find that 7= ♡. 16y^2-△=( 4y + 7)( 4y- 7) We are almost done! Since the squared terms a^2 and b^2 in the left hand side are the squares of the terms inside the binomials, we can write the the values of a and b correspondingly. a^2- b^2=( a+ b)( a- b) ⇓ ( 4y)^2- 7^2=( 4y+ 7)( 4y- 7) Great! Now, we can complete our equation to have it in the given form. 16y^2-△=( ◊ + 7)( 4y- ♡) ⇓ 16y^2-49=( 4y+ 7)( 4y- 7) Let's finally match our all missing values. △=49, ◊=4y, and ♡=7 Therefore the answer is option B.

This time we will start by recalling the formula to factor a perfect square trinomial. a^2+2 a b+ b^2=( a + b)^2 With this in mind, let's now rewrite and compare our equation. a^2+2 a b+ b^2=( a + b)^2 ⇓ ( 3x)^2+△+ 5^2=( ◊ + ♡)^2 As we can see, we can find the values of a and b such that a= 3x and b= 5. Since we can match a with ◊ and also b with ♡, we can substitute that ◊= 3x and ♡= 5. a^2+2 a b+ b^2=( a + b)^2 ⇓ ( 3x)^2+△+ 5^2=( 3x + 5)^2 Now, we will find the value of △ by substituting a= 3x and b= 5 into the middle term 2 a b.

Finally, we can write our all missing values. △=30x, ◊=3x, and ♡=5 Therefore, the answer is option C.

Factor the following trinomial.

c^{2} + 10 c + 25

We want to factor the given trinomial. c^2+10c+25 How do we know that the expression is a perfect square trinomial? Well, let's ask a few questions.

Is the first term a perfect square?	c^2= c^2 ✓
Is the last term a perfect square?	25= 5^2 ✓
Is the middle term twice the product of 5 and c?	10c=2* 5* c ✓

As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note there is an addition sign in the middle.

Factor the following expression completely.

3 k^{4} - 3

To factor the given expression, we will first identify and factor out the greatest common factor. We will then use the formula for the difference of two squares to finish the problem.

Factor Out the GCF

The greatest common factor (GCF) of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. The GCF of the given expression is 3.

Difference of Squares

In this part of the solution, we will have a look closely at the expression k^4-1. It can be expressed as the difference of two perfect squares.

Recall the formula to factor a difference of squares. a^2- b^2 ⇔ ( a+ b)( a- b) We can apply this formula to our expression. 3 ( ( k^2)^2- 1^2 ) ⇔ 3( k^2+ 1)( k^2- 1) Next, let's take a look at the expression k^2-1. Once again, we can express it as a difference of two perfect squares. Let's do it!

Now, we can apply the formula to factor our expression completely. 3 (k^2+1)( k^2- 1^2) ⇕ 3(k^2+1)( k+ 1)( k- 1)

Solve the following equation.

h^{2} - \frac{8}{9} h + \frac{1 6}{8 1} = 0

We will solve the given equation by factoring. To do so, we will start by identifying its first term and last term to check whether it is a perfect square trinomial or not. h^2-8/9h+16/81=0 ⇕ h^2-( 8/9 )h+ 16/81=0 Notice that the first term h^2 and the last term 1681 are the perfect squares. Let's see whether the middle term can be written as 2 a b to be able to say it is a perfect square trinomial. h^2-( 8/9 )h+ 16/81=0 ⇕ h^2-2 * h * ( 4/9 )+( 4/9 )^2=0 Great! As it is seen, the given trinomial can be rewritten in the form a^2-2 a b+ b^2. Therefore, it is a perfect square trinomial. Let's now factor it by applying the formula.

Now we are ready to use the Zero Product Property.

We found that the solution to the given equation is h= 49.

Factoring Quadratic Expressions Using Special Products

Catch-Up and Review

Showing Equivalent Expressions By Factoring

Perfect Square Trinomials and Factoring Them

Perfect Square Trinomial

Factoring a Perfect Square Trinomial

Framed Paintings

Hint

Solution

Factoring a Quadratic Expression and Analyzing Errors

Answer

Hint

Solution

Factoring an Expression as a Difference of Two Squares

Factoring a Difference of Two Squares

Extra

Writing Expressions for the Dimensions of a Rectangular Prism

Answer

Hint

Solution

Finding the Dimensions of a Rectangular Room

Hint

Solution

Deciding Which Method to Use for Factoring Expressions

Answer

Hint

Solution

Combining the Factorization of Special Products

Answer

Hint

Solution

Factor Out the GCF

Difference of Squares

Factoring Quadratic Expressions Using Special Products

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