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**CBSE Guide NCERT
Solutions of Class 9 Mathematics **

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**Chapter
****2, POLYNOMIALS**

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**NCERT Solutions of Math Exercise
2.4 **

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

**Solution:**

*(i) To have (x + 1) a factor of p(x) = x*

^{3 }+ x^{2}+ x + 1, substituting x = –1 must give p(−1) = 0, 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*

*Therefore, x + 1 is a factor of this polynomial.*

*(ii) For (x + 1) to be a factor of p(x) = x*

^{4}+ x^{3}+ x^{2}+ x + 1, p (−1) must be zero. Substituting x = –1 in the given polynomial we get,*p*

*(−1) = (−1)*

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

*As p(− 1) ≠ 0, (x + 1) is not a factor of this polynomial.*

*(iii)*Taking cue from the above try yourself.

*Question.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) To have g(x) = x + 1 as a factor of the given polynomial p(x), p(−1) must be zero. (*

*⇒*

*x = –1)*

*Substituting x = –1 in*

*2x*

^{3}+ x^{2}− 2x − 1*p*

*(−1) = 2(−1)*

^{3}+ (−1)^{2}− 2(−1) − 1*= 2(−1) + 1 + 2 − 1 = 0*

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

*(ii) p(x) = x*

^{3}+3x^{2}+ 3x + 1*g(x) = x + 2*

*⇒*

*x = –2, substituting this in the given polynomial p(x),*

*p*

*(−2) = (−2)*

^{3}+ 3(−2)^{2}+ 3(−2) + 1*= − 8 + 12 − 6 + 1*

*= −1*

*Since p(−2) ≠ 0,therefore, g(x) = x + 2 is not a factor of the given polynomial.*

*(iii)*(Taking hint from the above try yourself).

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

**Solution:**

*As it is given that (x – 1) is a factor of p(x), so after substituting x = 1 in p(x) we can solve for k by making p(1) = 0 in each case.*

*Question.4:**Factorize*

*(i) 12x*

^{2}− 7x + 1 (ii) 2x^{2}+ 7x + 3*(iii) 6x*

^{2}+ 5x − 6 (iv) 3x^{2}− x − 4**Solution:**

*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:**

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