Factoring can be done in the following ways.
Here we see, how we are factoring using algebraic identities.
a^{2} + 2ab + b^{2} = (a + b)^{2} ==> (a + b) (a + b)
a^{2} - 2ab + b^{2} = (a - b)^{2} ==> (a - b) (a - b)
a^{2} - b^{2} ==> (a + b) (a - b)
Factor each completely.
Problem 1 :
16n^{2} - 9
Solution:
The polynomial is a difference of two squares.
= (4n)^{2} - 3^{2}
= (4n + 3) (4n - 3)
Problem 2 :
4m^{2} - 25
Solution:
The polynomial is a difference of two squares.
= (2m)^{2} - 5^{2}
= (2m + 5) (2m -5)
Problem 3 :
16b^{2} - 40b + 25
Solution:
Since the first terms and last terms are perfect squares. The trinomial is a perfect square.
= 16b^{2} - 40b + 25
a = 4b, b = 5
Write the trinomial as a^{2} - 2ab + b^{2}
= (4b)^{2} - 2(4b)(5) + 5^{2}
= (4b - 5)^{2}
So, the linear factors are (4b - 5)(4b - 5).
Problem 4 :
4x^{2} - 4x + 1
Solution:
The trinomial is a perfect square. Factor
= 4x^{2} - 4x + 1
a = 2x, b = 1
Write the trinomial as a^{2} - 2ab + b^{2}
= (2x)^{2} - 2(2x)(1) + 1^{2}
= (2x - 1)^{2}
So, the linear factors are (2x - 1)(2x - 1).
Problem 5 :
9x^{2} - 1
Solution:
The polynomial is a difference of two squares.
= (3x)^{2} - 1^{2}
= (3x + 1) (3x - 1)
Problem 6 :
n^{2} - 25
Solution:
The polynomial is a difference of two squares.
= n^{2} - 5^{2}
= (n + 5) (n - 5)
Problem 7 :
n^{4} - 100
Solution:
The polynomial is a difference of two squares.
= (n^{2})^{2} - (10)^{2}
= (n^{2} + 10) (n^{2} - 10)
Problem 8 :
a^{4} - 9
Solution:
The polynomial is a difference of two squares.
= (a^{2})^{2} - (3)^{2}
= (a^{2} + 3) (a^{2} - 3)
Problem 9 :
k^{4} - 36
Solution:
The polynomial is a difference of two squares.
= (k^{2})^{2} - (6)^{2}
= (k^{2} + 6) (k^{2} - 6)
Problem 10 :
n^{4} - 49
Solution:
The polynomial is a difference of two squares.
= (n^{2})^{2} - (7)^{2}
= (n^{2} + 7) (n^{2} - 7)
Problem 11 :
98n^{2} - 200
Solution:
The polynomial is a difference of two squares.
= 98n^{2} - 200
= 2(49n^{2} - 100)
= 2((7n)^{2} - (10)^{2})
= 2(7n + 10) (7n - 10)
Problem 12 :
3 + 6b + 3b^{2}
Solution:
= 3 + 6b + 3b^{2}
= 3(1 + 2b + b²)
= 3(1 + b)²
So, the linear factors are 3 (1 + b)(1 + b).
Problem 13 :
400 - 36v^{2}
Solution:
= 400 - 36v²
= 4(100 - 9v²)
= 4(10)^{2} - (3v)^{2}
= 4(10 + 3v) (10 - 3v)
Problem 14 :
100x^{2} + 180x + 81
Solution:
The trinomial is a perfect square. Factor
= 100x^{2} + 180x + 81
a = 10x, b = 9
Write the trinomial as a^{2} + 2ab + b^{2}
= (10x)^{2} + 2(10x)(9) + 9^{2}
= (10x + 9)^{2}
So, the linear factors are (10x + 9)(10x + 9).
Problem 15 :
10n^{2} + 100n + 250
Solution:
The trinomial is a perfect square. Factor
= 10n^{2} + 100n + 250
= 10(n^{2} + 10n + 25)
a = n, b = 5
Write the trinomial as a^{2} + 2ab + b2
= 10((n)^{2} + 2(n)(5) + 5^{2})
= 10(n + 5)^{2}
So, the linear factors are 10(n + 5)(n + 5).
Problem 16 :
49n^{2} - 56n + 16
Solution:
The trinomial is a perfect square. Factor
= 49n^{2} - 56n + 16
a = 7n, b = 4
Write the trinomial as a2 - 2ab + b2
= (7n)^{2} - 2(7n)(4) + 4^{2}
= (7n + 4) (7n - 4)
Problem 17 :
49x^{2} - 100
Solution:
The polynomial is a difference of two squares.
= (7x)^{2} - (10)^{2}
= (7x + 10) (7x - 10)
Problem 18 :
1 - r^{2}
Solution:
The polynomial is a difference of two squares.
= 1^{2} - r^{2}
= (1 + r) (1 - r)
Problem 19 :
10p^{3} - 1960p
Solution:
= 10p^{3} - 1960p
= 10p(p2 - 196)
= 10p((p)^{2} - (14)^{2})
= 10p(p + 14) (p - 14)
Problem 20 :
343b^{2} - 7b^{4}
Solution:
= 343b² - 7b^{4}
= 7b^{2} (7^{2} - b^{2})
= 7b^{2}(7 + b) (7 - b)
Problem 21 :
81v^{4} - 900v^{2}
Solution:
= 81v^{4} - 900v^{2}
= 9v^{2}(3v^{2} - 10^{2})
= 9v^{2}(3v + 10) (3v - 10)
Problem 22 :
200m^{4} + 80m^{3} + 8m^{2}
Solution:
= 200m^{4} + 80m^{3} + 8m^{2}
= 8m²(25m² + 10m + 1)
a = 5m, b = 1
Write the trinomial as a^{2} + 2ab + b2
= 8m^{2}(5m)^{2} + 2(5m)(1) + 1^{2}
= 8m^{2}(5m + 1)^{2}
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM