Question

Find p(0), p(1) and p(2) for each of the following polynomials:
(i) P(y) = y^2 – y+1
(ii) p(t)=2+t+2t^2-t^3
(i) p(x)= x^3
(iv) p(x)=(x - 1) (x + 1)​

Answer 1

Concept :

For calculating p(0) , p(1) and p(2) of any of the given polynomial we just need to substitute 0 , 1 oR 2 in place of the variable in a polynomial. Doing so and proceeding with simple calculation we will end up getting our required answers.

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Question 1 :

⠀⠀⠀⌬ p(y) = y² - y + 1

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Solution :

$\sf\:p(y)=y^{2}-y+1$

Calculating p(0) by plugging 0 in place of y -

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

$\sf\leadsto\:p(0)=0-0+1$

$\sf\leadsto\:p(0)=\cancel{0}-\cancel{0}+1$

$\small{\underline{\boxed{\mathrm{\leadsto\:p(0)=1}}}}$ $\sf\color{purple}{⋆}$

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Calculating p(1) by plugging 1 in place of y -

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

$\sf\leadsto\:p(1)=1-1+1$

$\sf\leadsto\:p(1)=\cancel{1}-\cancel{1}+1$

$\small{\underline{\boxed{\mathrm{\leadsto\:p(1)=1}}}}$ $\sf\color{purple}{⋆}$

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Calculating p(2) by plugging 2 in place of y -

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

$\sf\leadsto\:p(2)=4-2+1$

$\sf\leadsto\:p(2)=2+1$

$\small{\underline{\boxed{\mathrm{\leadsto\:p(2)=3}}}}$ $\sf\color{purple}{⋆}$

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Question 2 :‎

⠀⠀⠀⌬ p(t) = 2 + t + 2t² - t³

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Solution :

$\sf\:p(t)=2+t+2t^{2}-t^{3}$

Calculating p(0) by plugging 0 in place of t -

$\sf\leadsto\:p(0)=2+0+2 \times 0^{2}-0^{3}$

$\sf\leadsto\:p(0)=2+0+2 \times 0-0$

$\sf\leadsto\:p(0)=2+0-0$

$\sf\leadsto\:p(0)=2+\cancel{0}-\cancel{0}$

$\small{\underline{\boxed{\mathrm{\leadsto\:p(0)=2}}}}$ $\sf\color{purple}{⋆}$

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Calculating p(1) by plugging 1 in place of t -

$\sf\leadsto\:p(1)=2+1+2 \times 1^{2}-1^{3}$

$\sf\leadsto\:p(1)=2+1+2 \times 1-1$

$\sf\leadsto\:p(1)=2+1+2-1$

$\sf\leadsto\:p(1)=3+2-1$

$\sf\leadsto\:p(1)=5-1$

$\small{\underline{\boxed{\mathrm{\leadsto\:p(1)=4}}}}$ $\sf\color{purple}{⋆}$

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Calculating p(2) by plugging 2 in place of t -

$\sf\leadsto\:p(2)=2+2+2 \times 2^{2}-2^{3}$

$\sf\leadsto\:p(2)=2+2+2 \times 4-8$

$\sf\leadsto\:p(2)=4+8-8$

$\sf\leadsto\:p(2)=12-8$

$\small{\underline{\boxed{\mathrm{\leadsto\:p(2)=4}}}}$ $\sf\color{purple}{⋆}$

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Question 3 :‎

⠀⠀⠀⌬ p(x) = x³

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Solution :

$\sf\:p(x)=x^{3}$

Calculating p(0) by plugging 0 in place of x -

$\sf\leadsto\:p(0)=0^{3}$

$\small{\underline{\boxed{\mathrm{\leadsto\:p(0)=0}}}}$ $\sf\color{purple}{⋆}$

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Calculating p(1) by plugging 1 in place of x -

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

$\small{\underline{\boxed{\mathrm{\leadsto\:p(1)=1}}}}$ $\sf\color{purple}{⋆}$

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Calculating p(2) by plugging 2 in place of x -

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

$\small{\underline{\boxed{\mathrm{\leadsto\:p(2)=8}}}}$ $\sf\color{purple}{⋆}$

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Question 4 :

⠀⠀⠀⌬ p(x) = ( x-1 ) ( x+1 )

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Solution :

$\sf\:p(x)=(x-1)(x+1)$

Calculating p(0) by plugging 0 in place of x -

$\sf\leadsto\:p(0)=(0-1)(0+1)$

$\sf\leadsto\:p(0)=(-1)(1)$

$\sf\leadsto\:p(0)=(-1) \times 1$

$\small{\underline{\boxed{\mathrm{\leadsto\:p(0)=(-1)}}}}$ $\sf\color{purple}{⋆}$

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Calculating p(1) by plugging 1 in place of x -

$\sf\leadsto\:p(1)=(1-1)(1+1)$

$\sf\leadsto\:p(1)=(0)(2)$

$\sf\leadsto\:p(1)=0 \times 2$

$\small{\underline{\boxed{\mathrm{\leadsto\:p(1)=0}}}}$ $\sf\color{purple}{⋆}$

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Calculating p(2) by plugging 2 in place of x -

$\sf\leadsto\:p(2)=(2-1)(2+1)$

$\sf\leadsto\:p(2)=(1)(3)$

$\sf\leadsto\:p(2)=1 \times 3$

$\small{\underline{\boxed{\mathrm{\leadsto\:p(2)=3}}}}$ $\sf\color{purple}{⋆}$

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