Question

2. Find p(0),p(1) and p(2) for each of the following polynomials.
( p(x) = x2-x +1
(ii) p(y) = 2 + y + 2y​

Answer 1

Answer:

1) p(0)=0×2-0+1

=1

p(1)=1×2-1+1

=2

p(2)=2×2+1

=5

2) p(0)=2+0+2×0

=4

p(1)=2+1+2×1

=5

p(2)=2+2+2×2

=8

Answer 2

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

Find p(0),p(1) and p(2) for each of the following polynomials.

(i) p(x) = x2-x +1

(ii) p(y) = 2 + y + 2y

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf \huge {(i) p(x) = x2-x +1}$

$\sf \bf {1.\:p(0)}$

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

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

$\sf \bf \implies {p(0) = 0-0+1}$

$\sf \bf \implies {\boxed {\boxed {p(0) =1}}}$

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$\sf \bf {2.\:p(1)}$

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

$\sf \bf \implies {p(1) = {(1)}^{2}-(1) +1}$

$\sf \bf \implies {p(2) = 1-1+1}$

$\sf \bf \implies {p(2) = 2-1}$

$\sf \bf \implies {\boxed {\boxed {p(1) =1}}}$

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$\sf \bf {3.\:p(2)}$

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

$\sf \bf \implies {p(2) = {(2)}^{2}-(2) +1}$

$\sf \bf \implies {p(2) = 4-2+1}$

$\sf \bf \implies {p(2) = 5-2}$

$\sf \bf \implies {\boxed {\boxed {p(2) =3}}}$

__________________________________ __________________________________

$\sf \bf \huge {(ii) p(y) = 2 + y + 2y}$

$\sf \bf {1.\:p(0)}$

$\sf \bf \implies {p(y) = 2+y+2y}$

$\sf \bf \implies {p(0) = 2+0+2(0)}$

$\sf \bf \implies {p(0) = 2+0+0}$

$\sf \bf \implies {\boxed {\boxed {p(0) =2}}}$

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$\sf \bf {2.\:p(1)}$

$\sf \bf \implies {p(y) = 2+y+2y}$

$\sf \bf \implies {p(1) = 2+1+2(1)}$

$\sf \bf \implies {p(0) = 2+1+2}$

$\sf \bf \implies {\boxed {\boxed {p(0) =5}}}$

_______________________________________

$\sf \bf {3.\:p(2)}$

$\sf \bf \implies {p(y) = 2+y+2y}$

$\sf \bf \implies {p(2) = 2+2+2(2)}$

$\sf \bf \implies {p(0) = 2+2+4}$

$\sf \bf \implies {\boxed {\boxed {p(0) =8}}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\sf \bf {Identities}$

$\sf \bf {{(a+b)}^{2}={a}^{2}+2ab+{b}^{2}}$

$\sf \bf {{(a-b)}^{2}={a}^{2}-2ab+{b}^{2}}$

$\sf \bf {(a+b) (a-b) ={a}^{2}-{b}^{2}}$

${\mathbb{\colorbox {orange} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {lime} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {aqua} {@suraj5070}}}}}}}}}}}}}}}$