Factoring can be done in the following ways.
Here we see, how we are factoring using algebraic identities.
a2 + 2ab + b2 = (a + b)2 ==> (a + b) (a + b)
a2 - 2ab + b2 = (a - b)2 ==> (a - b) (a - b)
a2 - b2 ==> (a + b) (a - b)
Factor each completely.
Problem 1 :
16n2 - 9
Solution:
The polynomial is a difference of two squares.
= (4n)2 - 32
= (4n + 3) (4n - 3)
Problem 2 :
4m2 - 25
Solution:
The polynomial is a difference of two squares.
= (2m)2 - 52
= (2m + 5) (2m -5)
Problem 3 :
16b2 - 40b + 25
Solution:
Since the first terms and last terms are perfect squares. The trinomial is a perfect square.
= 16b2 - 40b + 25
a = 4b, b = 5
Write the trinomial as a2 - 2ab + b2
= (4b)2 - 2(4b)(5) + 52
= (4b - 5)2
So, the linear factors are (4b - 5)(4b - 5).
Problem 4 :
4x2 - 4x + 1
Solution:
The trinomial is a perfect square. Factor
= 4x2 - 4x + 1
a = 2x, b = 1
Write the trinomial as a2 - 2ab + b2
= (2x)2 - 2(2x)(1) + 12
= (2x - 1)2
So, the linear factors are (2x - 1)(2x - 1).
Problem 5 :
9x2 - 1
Solution:
The polynomial is a difference of two squares.
= (3x)2 - 12
= (3x + 1) (3x - 1)
Problem 6 :
n2 - 25
Solution:
The polynomial is a difference of two squares.
= n2 - 52
= (n + 5) (n - 5)
Problem 7 :
n4 - 100
Solution:
The polynomial is a difference of two squares.
= (n2)2 - (10)2
= (n2 + 10) (n2 - 10)
Problem 8 :
a4 - 9
Solution:
The polynomial is a difference of two squares.
= (a2)2 - (3)2
= (a2 + 3) (a2 - 3)
Problem 9 :
k4 - 36
Solution:
The polynomial is a difference of two squares.
= (k2)2 - (6)2
= (k2 + 6) (k2 - 6)
Problem 10 :
n4 - 49
Solution:
The polynomial is a difference of two squares.
= (n2)2 - (7)2
= (n2 + 7) (n2 - 7)
Problem 11 :
98n2 - 200
Solution:
The polynomial is a difference of two squares.
= 98n2 - 200
= 2(49n2 - 100)
= 2((7n)2 - (10)2)
= 2(7n + 10) (7n - 10)
Problem 12 :
3 + 6b + 3b2
Solution:
= 3 + 6b + 3b2
= 3(1 + 2b + b²)
= 3(1 + b)²
So, the linear factors are 3 (1 + b)(1 + b).
Problem 13 :
400 - 36v2
Solution:
= 400 - 36v²
= 4(100 - 9v²)
= 4(10)2 - (3v)2
= 4(10 + 3v) (10 - 3v)
Problem 14 :
100x2 + 180x + 81
Solution:
The trinomial is a perfect square. Factor
= 100x2 + 180x + 81
a = 10x, b = 9
Write the trinomial as a2 + 2ab + b2
= (10x)2 + 2(10x)(9) + 92
= (10x + 9)2
So, the linear factors are (10x + 9)(10x + 9).
Problem 15 :
10n2 + 100n + 250
Solution:
The trinomial is a perfect square. Factor
= 10n2 + 100n + 250
= 10(n2 + 10n + 25)
a = n, b = 5
Write the trinomial as a2 + 2ab + b2
= 10((n)2 + 2(n)(5) + 52)
= 10(n + 5)2
So, the linear factors are 10(n + 5)(n + 5).
Problem 16 :
49n2 - 56n + 16
Solution:
The trinomial is a perfect square. Factor
= 49n2 - 56n + 16
a = 7n, b = 4
Write the trinomial as a2 - 2ab + b2
= (7n)2 - 2(7n)(4) + 42
= (7n + 4) (7n - 4)
Problem 17 :
49x2 - 100
Solution:
The polynomial is a difference of two squares.
= (7x)2 - (10)2
= (7x + 10) (7x - 10)
Problem 18 :
1 - r2
Solution:
The polynomial is a difference of two squares.
= 12 - r2
= (1 + r) (1 - r)
Problem 19 :
10p3 - 1960p
Solution:
= 10p3 - 1960p
= 10p(p2 - 196)
= 10p((p)2 - (14)2)
= 10p(p + 14) (p - 14)
Problem 20 :
343b2 - 7b4
Solution:
= 343b² - 7b4
= 7b2 (72 - b2)
= 7b2(7 + b) (7 - b)
Problem 21 :
81v4 - 900v2
Solution:
= 81v4 - 900v2
= 9v2(3v2 - 102)
= 9v2(3v + 10) (3v - 10)
Problem 22 :
200m4 + 80m3 + 8m2
Solution:
= 200m4 + 80m3 + 8m2
= 8m²(25m² + 10m + 1)
a = 5m, b = 1
Write the trinomial as a2 + 2ab + b2
= 8m2(5m)2 + 2(5m)(1) + 12
= 8m2(5m + 1)2
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM