Question

if p(x)=2x^3+3x^2+5x-7 then find the value of p(x)+p(-x)

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\sf{p(x) = 2 {x}^{3} + 3 {x}^{2} + 5x - 7}$

TO DETERMINE

$\sf{p(x) + p( - x) \: }$

CALCULATION

Here

$\sf{p(x) = 2 {x}^{3} + 3 {x}^{2} + 5x - 7}$

So

$\sf{p( - x) = 2 {( - x)}^{3} + 3 {( - x)}^{2} + 5 ( - x) - 7}$

$\therefore \: \: \sf{p( - x) = - 2 {x}^{3} + 3 {x}^{2} - 5x - 7}$

Hence

$\sf{p(x) + p( - x) \: }$

$= \sf{ 2 {x}^{3} + 3 {x}^{2} + 5x - 7 - 2 {x}^{3} + 3 {x}^{2} - 5x - 7\: }$

$= \sf{ 6 {x}^{2} - 14 \: }$

RESULT

$\boxed{ \sf{ \: \: p(x) + p( - x) \: = 6 {x}^{2} - 14\: }\: \: }$

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Show that the square of any positive integer can't be of the form 5m+2or 5m+3 where 'm' is a whole number

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Answer 2

$Given \: p(x) = p(x)=2x^3+3x^2+5x-7$

$\red{ Value \: of \: p(x)+p(-x)}$

$= (2x^3+3x^2+5x-7 )+[2(-x)^{3}+3(-x)^{2}+5(-x)-7]$

$= 2x^3+3x^2+5x-7 -2x^{3} +3x^{2} -5x-7$

$= 2x^{3} - 2x^{3} + 3x^{2} + 3x^{2} +5x -5x - 7 - 7$

$= 6x^{2} - 14$

Therefore.,

$\red{ Value \:of \: p(x) + p(-x) } \green {= 6x^{2} -14}$

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