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# Exercise 2.4 Class 9 maths 2. Polynomials - ncert solutions - Toppers Study

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## Exercise 2.4 Class 9 maths 2. Polynomials - ncert solutions - Toppers Study

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### Exercise 2.4 class 9 Mathematics Chapter 2. Polynomials

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## 2. Polynomials

### | Exercise 2.4 |

## Exercise 2.4 Class 9 maths 2. Polynomials - ncert solutions - Toppers Study

**Chapter 2. Polynomials**

**Exercise 2.4**

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

(i) x^{3 }+x^{2 }+ x +1 (ii) x^{4 }+ x^{3 }+ x^{2 }+ x +1

(iii) x^{4 }+ 3x^{3 }+ 3x^{2 }+ x +1

**Solution:**

(i) If (x + 1) is a factor of p(x) = x^{3}+ x^{2}+ x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).

P(x) = x^{3 }+ x^{2 }+ x + 1

P(−1)= (−1)^{3 }+ (−1)^{2 }+ (−1) + 1

= − 1 + 1 − 1 − 1

= 0

∵ P(x) = 0

Hence, x + 1 is a factor of this polynomial.

(ii) If (x + 1) is a factor of p(x) = x^{4}+ x^{3}+ x^{2}+ x + 1, then p (−1) must be zero, Otherwise (x + 1) is not a factor of p(x).

P(x) = x^{4}+ x^{3}+ x^{2}+ x + 1

P(−1) = (−1)^{4 }+ (−1)^{3 }+ (−1)^{2 }+ (−1) + 1

= 1 − 1 + 1 −1 + 1

= 1

As P(x) ≠ 0, (− 1)

Therefore, x + 1 is not a factor of this polynomial.

(iii) If (x + 1) is a factor of polynomial p(x) = x^{4}+ 3x^{3}+ 3x^{2}+ x + 1, then p (−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.

P(x) = x^{4}+ 3x^{3}+ 3x^{2}+ x + 1

P(−1) = (−1)^{4}+ 3(−1)^{3}+ 3(−1)^{2}+ (−1) + 1

= 1 − 3 + 3 − 1 + 1

= 1

As P(x) ≠ 0, (−1)

Therefore (x+1) is not a factor of this polynomial .

**Q.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) = 2x^{3}+ x^{2}− 2x − 1, g(x) = x + 1

(ii) P(x) = x^{3 }+ 3x^{2 }+ 3x + 1, g(x) = x + 2

(iii) P(x) = x^{3 }− 4 x^{2 }+ x + 6, g(x) = x − 3

**Solution:**

(i) If g(x) = x + 1 is a factor of the given polynomial p(x), then p (−1) must be zero.

P (x) = 2x^{3 }+ x^{2 }− 2x − 1

P (−1) = 2(−1)^{3}+ (−1)^{2}− 2(−1) − 1

= 2(−1) + 1 + 2 – 1

= 0

∵ P(x) = 0

Hence, g(x) = x + 1 is a factor of the given polynomial.

(ii) If g(x) = x + 2 is a factor of the given polynomial p(x), then p (−2) must be 0.

P (x) = x^{3}+3x^{2}+ 3x + 1

P (−2) = (−2)^{3}+ 3(−2)^{2}+ 3(−2) + 1

= − 8 + 12 − 6 + 1

= −1

As P(x) ≠ 0,

Hence, g(x) = x + 2 is not a factor of the given polynomial.

(iii) If g(x) = x − 3 is a factor of the given polynomial p(x), then P(3) must be 0.

P(x) = x^{3}− 4 x^{2}+ x + 6

P(3) = (3)^{3 }− 4(3)^{2 }+ 3 + 6

= 27 −36 + 9

= 0

Hence, g(x) = x − 3 is a factor of the given polynomial.

**Solution:**

: If x − 1 is a factor of polynomial p(x), then P(1) must be 0.

(i) P(x) = x^{2}+ x + k

P(1) = (1)^{2}+ 1 + k

= 1+1+

= 2+k

k =−2

(iv) P(x) = kx^{2}-3x + k

P(1) = k(1)^{2}-3(1) + k

= k-3+k

2k = 3

K = 3/2

**Question 4: Factorise: **

(i) 12x^{2}− 7x + 1 (ii) 2x^{2}+ 7x + 3

(iii) 6x^{2}+ 5x – 6 (iv) 3x^{2}− x − 4

**Solution:**

(i) 12x^{2}− 7x + 1 we can find two numbers,

Such that pq = 12 × 1 = 12 and p + q = −7.

They are p = −4 and q = −3

Here, 12x^{2}− 7x + 1

= 12x^{2}− 4x − 3x + 1

= 4x (3x − 1) − 1 (3x − 1)

= (3x − 1) (4x − 1)

(ii) 2x^{2}+ 7x + 3 we can find two numbers such that pq = 2 × 3= 6 and p + q = 7.

They are p = 6 and q = 1.

Here, 2x^{2 }+ 7x + 3

= 2x^{2}+ 6x + x + 3

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

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

(iii) 6x^{2}+ 5x − 6 we can find two numbers such that pq = −36 and p + q = 5.

They are p = 9 and q = −4.

Here, 6x^{2}+ 5x – 6

= 6x^{2}+ 9x − 4x – 6

= 3x (2x + 3) − 2 (2x + 3)

= (2x + 3) (3x − 2)

(iv) 3x^{2}− x − 4 we can find two numbers,

such that pq = 3 × (−4) = −12 and p + q = −1.

They are p = −4 and q = 3

Here, 3x^{2}− x − 4

= 3x^{2}− 4x + 3x – 4

= x (3x − 4) + 1 (3x − 4)

= (3x − 4) (x + 1)

**Question 5. Factorize:**

(i) x^{3}− 2x^{2}− x + 2 (ii) x^{3}+ 3x^{2}−9x − 5

(iii) x^{3}+ 13x^{2}+ 32x + 20 (iv) 2y^{3}+ y^{2}− 2y – 1

**Solution:**

(i) Let P(x) = x^{3}− 2x^{2}− x + 2 all the factor are there. These are ± 1, ± 2.

By trial method, P (1) = (1)^{3}− 2(1)^{2}− 1 + 2

= 1 − 2 − 1+ 2

= 0 Therefore, (x − 1) is factor of polynomial p(x)

Let us find the quotient on dividing x^{3}− 2x^{2}− x + 2 by x − 1.

By long division method

Now,

Dividend = Divisor × Quotient + remainder

x^{3}− 2x^{2}− x + 2 = (x – 1) ( X^{2}– x – 2) + 0

= (x – 1) (x^{2}–2x+x–2)

= (x – 1) [x (x–2) + 1(x–2)]

= (x – 1) (x + 1) (x – 2)

(ii) Let p(x) = x^{3 }– 3x^{2}−9x – 5 all the factor are there. These are ± 1, ± 2.

By trial method, p (–1) = (–1)^{3}– 3(1)^{2}− 9(1) – 5

= –1– 3–9–5 =0

Therefore (x+1) is the factor of polynomial p(x).

. Let us find the quotient on dividing x^{3}– 3x^{2}−9x – 5 by x+1.

By long division method

Now,

Dividend = Divisor × Quotient + remainder

x^{3}– 3x^{2}−9x – 5 = (x +1) ( X^{2}–4x – 5) + 0

=(x + 1) (x^{2}–5x+x–5)

=(x + 1) [x (x–5) +1(x–5)]

=(x + 1) (x + 1) (x – 5)

(iii) Let p(x) = x^{3}+ 13x^{2}+ 32x + 20 all the factor are there.

These are ± 1, ± 2, ± 3, ± 4.

By trial method, p (–1) = (–1)^{3}+13(–1)^{2}+ 32(–1) +20

= –1+13–32+20 = 0

Therefore (x+1) is the factor of polynomial p(x).

Let us find the quotient on dividing x^{3}+ 13x^{2}+ 32x + 20 by x+1

By long division method

Now,

Dividend = Divisor × Quotient + remainder

x^{3 }+13x^{2 }+ 32x + 20 = (x +1) ( x^{2} + 12x + 20) + 0

=(x + 1) (x^{2}+10x+2x+20)

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

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

(iv) Let p(y) = 2y^{3}+ y^{2}− 2y – 1 all the factor are there. These are ± 1, ± 2.

By trial method, p (1) =2(1)^{3 }+ (1)^{2 }– 2(1) – 1

=2 + 1 – 2 – 1 =0

Therefore (y–1) is the factor of polynomial p(y).

Let us find the quotient on dividing 2y^{3}+ y^{2}− 2y – 1 by y–1.

By long division method

Now,

Dividend = Divisor × Quotient + remainder

2y^{3}+ y^{2}− 2y −1 =(y − 1) (2y^{2}+3y + 1)

= (y − 1) (2y^{2}+2y

= (y − 1) [2y (y+1) + 1 (y + 1)]

= (y − 1) (y + 1) (2y + 1)

##### Other Pages of this Chapter: 2. Polynomials

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