Question

Find p(0),p(1),p(2) the given polynomial is p(y)=(x-1)²(x+1)²​

Answer 1

(i) p(y)=y

2

−y+1

p(0)=(0)

2

−(0)+1=0+0+1=1

p(1)=(1)

2

−(1)+1=1−1+1=1

p(2)=(2)

2

−(2)+1=4−2+1=3

(ii) p(t)=2+t+2t

2

−t

3

p(0)=2+(0)+2(0)

2

−t

3

=2+0+0−0=2

p(1)=2+(1)+2(2)

2

−(1)

3

=2+1+2−1=4

p(2)=2+(2)+2(2)

2

−(2)

3

=2+2+8−8=4

(iii) p(x)=x

3

p(0)=(0)

3

=0

p(1)=(1)

3

=1

p(2)=(2)

3

=8

(iv) p(x)=(x−1)(x+1)

p(0)=(0−1)(0+1)=(−1)(1)=−1

p(1)=(1−1)(1+1)=(0)(2)=0

p(2)=(2−1)(2+1)=(1)(3)=3

Answer 2

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$\: \: \bigstar\large\pink { \bold{ \underline{ \underline{correct \: answer}}}}$

$\: \: \large \blue{\bold{ \underline{ given}}} :$

$\bullet\bf \: find \: p(0) \: \: p(1) \: and \: p(2) \: the \: given \: polynomial \: is \: p(y) = ( {x - 1)}^{2} {(x + 1)}^{2}$

$\: \\ \quad \implies \displaystyle \bf{p(x) = {(x - 1)}^{2} {(x + 1)}^{2} }$

$\: \quad \implies \displaystyle \sf{p(0) = {(0 - 1)}^{2} ( {0 + 1)}^{2} }$

$\: \: \quad \implies \displaystyle \sf{( { - 1})^{2} {(1)}^{2} } = 1$

$\: \: \quad \implies \green{\bf{ \fbox{1}}}$

$\: \quad \mapsto \red{\displaystyle \bf{now \: put \: x = 1}}$

$\: \: \quad \mapsto \displaystyle \sf{p(1) = {(1 - 1)}^{2} {(1 + 1)}^{2} }$

$\: \: \quad \mapsto \displaystyle \sf{p(1) = {0}^{2} \times {2}^{2} }$

$\: \: \quad \mapsto \displaystyle \sf{p(1) = 0}$

$\: \: \quad \mapsto \displaystyle \sf{ \boxed{p(1) = 0}}$

$\: \quad \leadsto \red{ \displaystyle \bf{now \: put \: x = 2}}$

$\: \: \quad \leadsto \displaystyle \sf{p(2) = {(2 - 1)}^{2} {(2 + 1)}^{2} }$

$\: \: \quad \leadsto \displaystyle \sf{ p(2) = {(1)}^{2} {(3)}^{2} }$

$\: \: \quad \leadsto \displaystyle \sf{p(2) = 9}$

$\: \: \quad \leadsto \pink{\displaystyle \sf{p(2) = 9}}$