Question

if p(x)=x^3+3x^2-2x+4,then find the value of p(2)+p(-2)-p(0)​

Answer 1

SOLUTION

GIVEN

$\sf{p(x) = {x}^{3} + 3 {x}^{2} - 2x + 4}$

TO DETERMINE

$\sf{p(2) + p( - 2) - p(0)}$

EVALUATION

Here it is given that

$\sf{p(x) = {x}^{3} + 3 {x}^{2} - 2x + 4}$

Now

$\sf{p(2) = {(2)}^{3} + 3. {( 2)}^{2} - 2.2 + 4}$

$\sf{ = 8 + 12 - 4 + 4}$

$\sf{ = 20}$

Again

$\sf{p( - 2) = {( - 2)}^{3} + 3. {( - 2)}^{2} - 2.( - 2 )+ 4}$

$\sf{= - 8 + 12 + 4+ 4}$

$\sf{ = 12}$

Again

$\sf{p( 0) = {( 0)}^{3} + 3. {( 0)}^{2} - 2.( 0 )+ 4}$

$\sf{ = 0 + 0 - 0 + 4}$

$= 4$

$\sf{ \therefore \: \: p(2) + p( - 2) - p(0)}$

$= \sf{20 + 12 - 4}$

$\sf{ = 28}$

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Answer 2

Answer

p(2) + p(-2) - p(0) = 28

Step By Step Explanation

p(2) + p(-2) - p(0)

= [2³ + 3(2)² - 2(2) + 4] + [(-2)³ + 3(-2)² - 2(-2) + 4] - [0³ + 3(0)² - 2(0) + 4]

= [8 + 3(4) - 4 + 4] + [-8 + 3(4) - (-4) + 4] - [0 + 3(0) - 0 + 4]

= [8 + 12 - 4 + 4] + [-8 + 12 + 4 + 4] - [0 + 0 - 0 + 4]

= [8 + 12] + [-8 + 12 + 8] - [4]

= 20 + 12 - 4

= 32 - 4

= 28