Factorize: {Exercise 2.4, Polynomials, class 9}

 (i) 12x2–7x+1

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = -7 and product =1×12 = 12

We get -3 and -4 as the numbers [-3+-4=-7 and -3×-4 = 12]

12x2–7x+1= 12x2-4x-3x+1

= 4x(3x-1)-1(3x-1)

= (4x-1)(3x-1)

(ii) 2x2+7x+3

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = 7 and product = 2×3 = 6

We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 6]

2x2+7x+3 = 2x2+6x+1x+3

= 2x (x+3)+1(x+3)

= (2x+1)(x+3)

(iii) 6x2+5x-6 

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = 5 and product = 6×-6 = -36

We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -36]

6x2+5x-6 = 6x2+9x–4x–6

= 3x(2x+3)–2(2x+3)

= (2x+3)(3x–2)

(iv) 3x2–x–4 

Solution:

Using the splitting the middle term method,

We have to find a number whose sum = -1 and product = 3×-4 = -12

We get -4 and 3 as the numbers [-4+3 = -1 and -4×3 = -12]

3x2–x–4 = 3x2–4x+3x–4

= x(3x–4)+1(3x–4)

= (3x–4)(x+1)

(i) x3–2x2–x+2

Solution:

Let p(x) = x3–2x2–x+2

Factors of 2 are ±1 and ± 2

Now,

p(x) = x3–2x2–x+2

p(−1) = (−1)3–2(−1)2–(−1)+2

= −1−2+1+2

= 0

Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor × Quotient + Remainder

(x+1)(x2–3x+2) = (x+1)(x2–x–2x+2)

= (x+1)(x(x−1)−2(x−1))

= (x+1)(x−1)(x-2)

(ii) x3–3x2–9x–5

Solution:

Let p(x) = x3–3x2–9x–5

Factors of 5 are ±1 and ±5

By trial method, we find that

p(5) = 0

So, (x-5) is factor of p(x)

Now,

p(x) = x3–3x2–9x–5

p(5) = (5)3–3(5)2–9(5)–5

= 125−75−45−5

= 0

Therefore, (x-5) is the factor of  p(x)

Now, Dividend = Divisor × Quotient + Remainder

(x−5)(x2+2x+1) = (x−5)(x2+x+x+1)

= (x−5)(x(x+1)+1(x+1))

= (x−5)(x+1)(x+1)

(iii) x3+13x2+32x+20

Solution:

Let p(x) = x3+13x2+32x+20

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By trial method, we find that

p(-1) = 0

So, (x+1) is factor of p(x)

Now,

p(x)= x3+13x2+32x+20

p(-1) = (−1)3+13(−1)2+32(−1)+20

= −1+13−32+20

= 0

Therefore, (x+1) is the factor of p(x)

Now, Dividend = Divisor × Quotient +Remainder

(x+1)(x2+12x+20) = (x+1)(x2+2x+10x+20)

= (x+1)x(x+2)+10(x+2)

= (x+1)(x+2)(x+10)

(iv) 2y3+y2–2y–1

Solution:

Let p(y) = 2y3+y2–2y–1

Factors = 2×(−1)= -2 are ±1 and ±2

By trial method, we find that

p(1) = 0

So, (y-1) is factor of p(y)

Now,

p(y) = 2y3+y2–2y–1

p(1) = 2(1)3+(1)2–2(1)–1

= 2+1−2

= 0

Therefore, (y-1) is the factor of p(y)

Now, Dividend = Divisor × Quotient + Remainder

(y−1)(2y2+3y+1) = (y−1)(2y2+2y+y+1)

= (y−1)(2y(y+1)+1(y+1))

= (y−1)(2y+1)(y+1)