factorization

We know that, Factors of 12 2, 3, 4, 6. Prime factors of 12 3 . 2 . 2

Further factors of Factor 1 Further factors of Factor 2 What is factorization.? The process of expressing any polynomial as a product of it’s factors is called FACTORIZATION. Eg. 9x² = (3x) (3x) 2a²b + 2ab² = (2ab) (a + b) 3x² + x -6 = (3x - 3) (x + 2) Algebraic expression Factor 1 Factor 2 Further factors of Factor 1 Further factors of Factor 2 9x² (3x) 3x = 3 . x 2a²b + 2ab² (2ab) (a + b) 2ab = 2 . a. b 3x² + x – 6 (3x - 3) (x + 2)

Points to remember, A whole number greater than 1 for which the only factors are 1 and itself, is called a prime number. Eg. 2,3,4,5… A whole number greater than 1 which has more than 2 factors is called a composite number. Eg. 4,6,8,10… 1 is a factor of any number, 1 is neither prime nor composite. Every natural number other than 1 is neither prime nor composite. 2 is the only even prime number.

Factorization by Taking out the common factor Eg. 4x + 8 = 4x + (2 . 4) = 4 (x + 2) 4 is common to both terms. The factors of 4x + 8 = 4 and (x + 2) Factors 4 and (x + 2) are irreducible [Note: a factors that can’t be factorized further is known as irreducible factors]

Factorization by Grouping the terms… Eg. X³ - 2x² + x – 2 = x²(x -2) + 1(x - 2) [step 1: In first two terms x² is common - In second term 1 is common - lets take out those 2 term] = (x² + 1) (x – 2) [step 2: let’s group the common term from the both first and second term (i.e) (x – 2), then group the remaining terms in another group]

Factorization by using identities… “polynomial expression can be Recall, factorized by using IDENTITIES..” (x + y)² (x + y) (x + y) x² + 2xy +y² (x - y)² (x - y) (x - y) x² - 2xy +y² x² - y² (x + y) (x - y) - (x + a) (x + b) X2 + (a+b)x + ab

Examples… Factorize the following expression by using identities. 1. x² + 4x + 4 Solution, x² + 4x + 4 = x² + 2(2)x + 2 [compare with suitable Identity] = (x + 2)² = (x + 2) (x + 2) i.e, (a + b)² = a² + 2ab + b² = x² + 4x + 4 a = x, 2ab = 2x(2), b = 2

2. x² + 4x + 4 Solution, x² + 6x + 9 = x² - 2x(3) + 9 [compare with 2. x² + 4x + 4 Solution, x² + 6x + 9 = x² - 2x(3) + 9 [compare with suitable Identity] = (x - 3)² = (x - 3) (x - 3) i.e., (a-b)² = a² - 2ab + b² = x² - 2x(3) + 9 a = x, 2ab = 2x(3), b = 3

3. m4 – n4 solution: m4 – n4 = m4 – n4 [compare with suitable Identity (a2 – b 2)] = (m2)2 – (n 2)2 = (m2 + n2) (m2 – n2) = (m2 + n2) (m + n) (m – n) i.e, (a2 – b2) = (a - b) (a + b) m4 – n4 a2 = (m2)2 a = m2 b2 = (n2)2 b = n2

x2+(a + b)x + ab = (x + a) (x + b) 4. x² + 7x + 12 solution: x² + 7x + 12 = x² + 7x + 12 [compare with suitable Identity] = x² + (3 + 4)x + 12 = x² + 3x + 4x + 12 = (x² + 3x) + (4x + 12) = (x.x + 3.x) + (4.x +3.4) = x (x + 3) + 4 (x + 3) = (x + 3) (x + 4) x2+(a + b)x + ab = (x + a) (x + b) x² + 7x + 12 a + b = 7, ab = 12 a b a + b ab choice 1 6 7 2 5 10 3 4 12

x2+ (a + b)x + ab = (x + a) (x + b) 5. x² - 3x - 4 solution: x² - 3x - 4 = x² - 3x - 4 [compare with suitable identity] = x² + (1 + (-4))x + (1.(-4)) = x² + (1x + (-4)x) + (1.(-4)) = (x² + 1x) + ((-4)x + 1.(-4)) = (x.x + 1.x) + ((-4).x +1.(-4)) = x (x + 1) + (-4) (x + 1) = (x + 1) (x + (- 4)) = (x + 1) (x - 4) x2+ (a + b)x + ab = (x + a) (x + b) x² - 3x - 4 a + b = -3, ab = -4 a b a + b ab choice -1 3 2 -3 -2 -4 1

6. 4x²y - 7xy². Solution:. 4x²y - 7xy² = 4x²y - 7xy² 6. 4x²y - 7xy² Solution: 4x²y - 7xy² = 4x²y - 7xy² [separate the terms] = (4.x.x.y) – (7.x.y.y) [take out the common terms x.y] = xy (4x – 7y) [group the terms]

Try These Factorize the following: y2− 8y + 2y – 16 x2 +7x + 10