Question

If p(x) = x³-x²+x+1, then find the value of (1)+(−1)+(2)\2

Answer 1

Step-by-step explanation:

x^3 +x^2 +x +1 =0

=> x^2(x+1) + 1(x+1)=0

=> (x+1)(x^2 +1)=0, Therefore ,

Either x+1=0 or x^2+1=0, if x+1=0 => x=-1

If x^2 +1=0 => x^2 = -1 , x^2 = i^2 =>x= √(i^2)

x=±i , then the required solution

x= -1 , +i , -i

Answer 2

Given the polynomial p(x)=x

3

−x

2

+x+1

\text{we have to find the value of }\frac{1}{2}(p(-1)+p(1))we have to find the value of

2

1

(p(−1)+p(1))

p(x)=x^3-x^2+x+1p(x)=x

3

−x

2

+x+1

Put x=-1 and x=1

p(-1)=(-1)^3-(-1)^2+(-1)+1=-1-1-1+1=-2p(−1)=(−1)

3

−(−1)

2

+(−1)+1=−1−1−1+1=−2

p(1)=(1)^3-(1)^2+(1)+1=1-1+1+1=2p(1)=(1)

3

−(1)

2

+(1)+1=1−1+1+1=2

\frac{1}{2}(p(-1)+p(1))

2

1

(p(−1)+p(1))

=\frac{1}{2}(-2+2)=0=

2

1

(−2+2)=0

\text{Hence, the value is }\frac{1}{2}(p(-1)+p(1))\text{ is 0.}Hence, the value is

2

1

(p(−1)+p(1)) is 0.