Question

28. Ifp * (x) = x ^ 2 + 3x - 2 , then find the value of p(2) - p(- 2) + p * 1/2​

Answer 1

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• Given polynomial is p(x)=x
• Given polynomial is p(x)=x 2
• Given polynomial is p(x)=x 2 −3x+2
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we get
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0)
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we get
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we getp(1)=(1)
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we getp(1)=(1) 2
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we getp(1)=(1) 2 −3(1)+2
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we getp(1)=(1) 2 −3(1)+2⇒p(1)=1−3+2
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we getp(1)=(1) 2 −3(1)+2⇒p(1)=1−3+2⇒p(1)=0
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we getp(1)=(1) 2 −3(1)+2⇒p(1)=1−3+2⇒p(1)=0Replace x by 2, we get
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we getp(1)=(1) 2 −3(1)+2⇒p(1)=1−3+2⇒p(1)=0Replace x by 2, we getp(2)=(2)
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we getp(1)=(1) 2 −3(1)+2⇒p(1)=1−3+2⇒p(1)=0Replace x by 2, we getp(2)=(2) 2
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we getp(1)=(1) 2 −3(1)+2⇒p(1)=1−3+2⇒p(1)=0Replace x by 2, we getp(2)=(2) 2 −3(2)+2
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we getp(1)=(1) 2 −3(1)+2⇒p(1)=1−3+2⇒p(1)=0Replace x by 2, we getp(2)=(2) 2 −3(2)+2⇒p(2)=4−6+2
• Given polynomial is p(x)=x 2 −3x+2Replace x by 0, we getp(0)=(0) 2 −3(0)+2⇒p(0)=0+0+2⇒p(0)=2Replace x by 1, we getp(1)=(1) 2 −3(1)+2⇒p(1)=1−3+2⇒p(1)=0Replace x by 2, we getp(2)=(2) 2 −3(2)+2⇒p(2)=4−6+2⇒p(2)=0