A trinomial is a perfect square if :
• The first and last terms are perfect squares.
• The middle term is two times one factor from the first term and one factor from the last term.
a^{2} + 2ab + b^{2} = (a + b)(a + b) = (a + b)^{2}
a^{2} - 2ab + b^{2} = (a - b)(a - b) = (a - b)^{2}
Examples :
x^{2} + 6x + 9 = (x + 3)(x + 3) = (x + 3)^{2}
x^{2} -2x + 1 = (x - 1)(x - 1) = (x - 1)^{2}
Determine whether each trinomial is a perfect square. If so, factor. If not, explain.
Example 1 :
x^{2} + 12x + 36
The trinomial is a perfect square. Factor.
Method 1 : Factor.
x^{2} + 12x + 36
Factors of 36 1 and 36 2 and 18 3 and 12 4 and 9 6 and 6 |
Sum 37 ✗ 20 ✗ 15 ✗ 13 ✗ 36 ✓ |
= (x + 6)(x + 6)
= (x + 6)^{2}
Method 2 : Use the rule.
x^{2} + 12x + 36
a = x, b = 6.
Write the trinomial as a^{2} + 2ab + b^{2}.
= x^{2} + 2(x)(6) + 6^{2}
Write the trinomial as (a + b)^{2}.
= (x + 6)^{2}
Example 2 :
4x^{2} - 12x + 9
The trinomial is a perfect square. Factor.
= 4x^{2} - 12x + 9
a = 2x, b = 3.
Write the trinomial as a^{2} - 2ab + b^{2}.
= (2x)^{2} - 2(2x)(3) + 3^{2}
Write the trinomial as (a - b)^{2}.
= (2x - 3)^{2}
Example 3 :
x^{2} + 9x + 16
2(x · 4) ≠ 9x
x^{2} + 9x + 16 is not a perfect-square trinomial because
9x ≠ 2(x · 4)
a^{2} - b^{2} = (a + b)(a - b)
Example :
x^{2} - 9 = (x + 3)(x - 3)
The difference of two squares can be written as the product (a + b)(a - b) . You can use this pattern to factor some polynomials.
A polynomial is a difference of two squares if:
• There are two terms, one subtracted from the other.
• Both terms are perfect squares.
Determine whether each binomial is a difference of two squares. If so, factor. If not, explain.
Example 4 :
x^{2} - 81
The polynomial is a difference of two squares.
= x^{2} - 81
= x^{2} - 9^{2}
a = x and b = 9, write the polynomial as (a + b)(a - b).
= (x + 9)(x - 9)
Example 5 :
9p^{4} - 16q^{2}
The polynomial is a difference of two squares.
= 9p^{4} - 16q^{2}
= (3p^{2})^{2} - (4q)^{2}
a = 3p^{2} and b = 4q, write the polynomial as (a + b)(a - b).
= (3p^{2} + 4q)(3p^{2} - 4q)
Example 6 :
x^{6} - 7y^{2}
7y^{2} is not a perfect square.
x^{6} - 7y^{2 }is not the difference of two squares because 7y^{2 }is not a perfect square.
Example 7 :
The park in the center of the Place des Vosges in Paris, France, is in the shape of a square. The area of the park is (25x^{2} + 70x + 49) ft^{2} . The side length of the park is in the form cx + d, where c and d are whole numbers. Find an expression in terms of x for the perimeter of the park. Find the perimeter when x = 8 ft.
Understand the Problem :
The answer will be an expression for the perimeter of the park and the value of the expression when x = 8.
List the important information :
• The park is a square with area (25x^{2} + 70x + 49) ft^{2}.
• The side length of the park is in the form cx + d, where c and d are whole numbers.
Make a Plan :
The formula for the area of a square is area = (side) 2 .
Factor 25x^{2} + 70x + 49 to find the side length of the park. Write a formula for the perimeter of the park, and evaluate the expression for x = 8.
Solve :
25x^{2} + 70x + 49
a = 5x, b = 7.
Write the trinomial as a^{2} + 2ab + b^{2}.
= (5x)^{2} + 2(5x)(7) + 7^{2}
Write the trinomial as (a + b)^{2}.
= (5x + 7)^{2}
25x^{2} + 70x + 49 = (5x + 7)(5x + 7)
The side length of the park is (5x + 7) ft.
Write a formula for the perimeter of the park.
P = 4s
Substitute the side length for s.
= 4(5x + 7)
Distribute 4.
= 20x + 28
An expression for the perimeter of the park in feet is 20x + 28.
Evaluate the expression when x = 8.
P = 20x + 28
Substitute 8 for x.
= 20 (8) + 28
= 188
When x = 8 ft, the perimeter of the park is 188 ft.
Look Back :
For a square with a perimeter of 188 ft, the side length is
188/4 = 47 ft
and the area is
47^{2} = 2209 ft^{2}
Evaluate 25x^{2} + 70x + 49 for x = 8.
= 25(8)^{2} + 70(8) + 49
= 16001 + 560 + 49
= 2209 ✓
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