Question

"Question 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
(iv) x^3 - x^2 - (2 + 2^(1/2))x + 2^(1/2)

Class 9 - Math - Polynomials Page 43"

Answer 1

If (x+1) is a factor of given polynomial say p(X) , then at x= -1, p(x)  will become zero, otherwise it is not a factor of given polynomial.

 

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Solution:

 

(i) x³+x²+x+1

 

Let p(x)= x³+x²+x+1

 

 

The zero of x+1 is -1.

 

On putting x= -1

p(−1)=(−1)³+(−1)²+(−1)+1

=−1+1−1+1=0

Hence, by factor theorem, x+1 is a factor of x³+x²+x+1

 

(ii) x4 + x3 + x2 + x + 1

 

Let p(x)= x⁴+x³+x²+x+1

 

The zero of x+1 is -1.

 

On putting x= -1

p(−1)=(−1)⁴+(−1)³+(−1)²+(−1)+1

 

=1−1+1−1+1=1≠0

 

Hence, by factor theorem, x+1 is not  a factor

of x⁴+x³+x²+x+1

 

 (iii) x4 + 3x3 + 3x2 + x + 1 

 

Let p(x)= x⁴+3x³+3x²+x+1

 

The zero of x+1 is -1.

 

On putting x= -1

p(−1)=(−1)⁴+3(−1)³+3(−1)²+(−1)+1

 

=1−3+3−1+1

=1≠0

 

Hence,by factor theorem, x+1 is not a factor of x⁴+3x³+3x²+x+1

 

(iv) x³–x²–(2+√2)x+√2

 

Let p(x)= x³–x²–(2+√2)x+√2

 

The zero of x+1 is -1.

 

On Putting x= -1

p(−1)=(−1)³–(−1)²–(2+√2)(−1)+√2

Hence, by factor theorem, x+1 is not a factor of x³–x²–(2+√2)x+√2.

 

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Answer 2

hello friends
here is your answer



factor=(x+1)
so x+1=0
=>x=-1

(i) p(x)=x³+x²+x+1
p(-1)=(-1)³+(-1)²+(-1)+1
=-1+1-1+1
=0
so (x+1) is the factor of x³+x²+x+1

(ii) p(x)= x^4+x³+x²+x+1
p(-1)=(-1)^4+(-1)³+(-1)²+(-1)+1
=1-1+1-1+1
=1
so (x+1) is not the factor of x^4+x³+x²+x+1

(iii) p(x)=x^4+3x³+3x²+x+1
p(-1)=(-1)^4+3(-1)³+3(-1)²+(-1)+1
= 1-3+3-1+1
= 1
so (x+1) is not the factor of x^4+3x³+3x²+x+1

(iv) p(x)=x³-x²-(2+√2)x+√2
[ note i have changed 2^(1/2) to √2 because they are equal ]
p(-1)=(-1)³-(-1)²-(2+√2)(-1)+√2
= -1-1-(-2-√2)+√2
= -2+2+√2+√2
= 2√2
so x³-x²-(2+√2)x+√2 don't have (x+1) as a factor.


Hope that this will help you
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Thank you

-Sneha