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## Exercise 2.4 Page: 43

1. Determine which of the following polynomials has (x + 1) a factor:

(i) x3+x2+x+1

Solution:

Let p(x) = x3+x2+x+1

The zero of x+1 is -1. [x+1 = 0 means x = -1]

p(−1) = (−1)3+(−1)2+(−1)+1

= −1+1−1+1

= 0

∴By factor theorem, x+1 is a factor of x3+x2+x+1

(ii) x4+x3+x2+x+1

Solution:

Let p(x)= x4+x3+x2+x+1

The zero of x+1 is -1. . [x+1= 0 means x = -1]

p(−1) = (−1)4+(−1)3+(−1)2+(−1)+1

= 1−1+1−1+1

= 1 ≠ 0

∴By factor theorem, x+1 is not a factor of x4 + x3 + x2 + x + 1

(iii) x4+3x3+3x2+x+1

Solution:

Let p(x)= x4+3x3+3x2+x+1

The zero of x+1 is -1.

p(−1)=(−1)4+3(−1)3+3(−1)2+(−1)+1

=1−3+3−1+1

=1 ≠ 0

∴By factor theorem, x+1 is not a factor of x4+3x3+3x2+x+1

(iv) x– x2– (2+√2)x +√2

Solution:

Let p(x) = x3–x2–(2+√2)x +√2

The zero of x+1 is -1.

p(−1) = (-1)3–(-1)2–(2+√2)(-1) + √2 = −1−1+2+√2+√2

= 2√2 ≠ 0

∴By factor theorem, x+1 is not a factor of x3–x2–(2+√2)x +√2

2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p(x) = 2x3+x2–2x–1, g(x) = x+1

Solution:

p(x) = 2x3+x2–2x–1, g(x) = x+1

g(x) = 0

⇒ x+1 = 0

⇒ x = −1

∴Zero of g(x) is -1.

Now,

p(−1) = 2(−1)3+(−1)2–2(−1)–1

= −2+1+2−1

= 0

∴By factor theorem, g(x) is a factor of p(x).

(ii) p(x)=x3+3x2+3x+1, g(x) = x+2

Solution:

p(x) = x3+3x2+3x+1, g(x) = x+2

g(x) = 0

⇒ x+2 = 0

⇒ x = −2

∴ Zero of g(x) is -2.

Now,

p(−2) = (−2)3+3(−2)2+3(−2)+1

= −8+12−6+1

= −1 ≠ 0

∴By factor theorem, g(x) is not a factor of p(x).

(iii) p(x)=x3–4x2+x+6, g(x) = x–3

Solution:

p(x) = x3–4x2+x+6, g(x) = x -3

g(x) = 0

⇒ x−3 = 0

⇒ x = 3

∴ Zero of g(x) is 3.

Now,

p(3) = (3)3−4(3)2+(3)+6

= 27−36+3+6

= 0

∴By factor theorem, g(x) is a factor of p(x).

3. Find the value of k, if x–1 is a factor of p(x) in each of the following cases:

(i) p(x) = x2+x+k

Solution:

If x-1 is a factor of p(x), then p(1) = 0

By Factor Theorem

⇒ (1)2+(1)+k = 0

⇒ 1+1+k = 0

⇒ 2+k = 0

⇒ k = −2

(ii) p(x) = 2x2+kx+√2

Solution:

If x-1 is a factor of p(x), then p(1)=0

⇒ 2(1)2+k(1)+√2 = 0

⇒ 2+k+√2 = 0

⇒ k = −(2+√2)

(iii) p(x) = kx22x+1

Solution:

If x-1 is a factor of p(x), then p(1)=0

By Factor Theorem

⇒ k(1)2-√2(1)+1=0

⇒ k = √2-1

(iv) p(x)=kx2–3x+k

Solution:

If x-1 is a factor of p(x), then p(1) = 0

By Factor Theorem

⇒ k(1)2–3(1)+k = 0

⇒ k−3+k = 0

⇒ 2k−3 = 0

⇒ k= 3/2

4. Factorize:

(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–x–4

= 3x2–4x+3x–4

= x(3x–4)+1(3x–4)

= (3x–4)(x+1)

5. Factorize:

(i) x3–2x2–x+2

Solution:

Let p(x) = x3–2x2–x+2

Factors of 2 are ±1 and ± 2

By trial method, we find that

p(1) = 0

So, (x+1) is factor of p(x)

Now,

p(x) = x3–2x2–x+2

p(−1) = (−1)3–2(−1)2–(−1)+2

= −1−1+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−5)x(x+2)+10(x+2)

= (x−5)(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)

Class 9th Maths Polynomial