Let p(x) be a fourth degree polynomial with a leading coefficent 2 such that p(-2) = 34, p(-1) = 10, p(1) = 10, and p(2) = 34. Find p(0)???

2 Answers
Jun 8, 2017

P(0)=10

Explanation:

P(x)=2x^4+bx^3+cx^2+dx+e

we want P(0)

P(0)=0+0+0+0+e=e

so we only need to find the value for e

To do that we will use the information given.

P(2)=2xx2^4+8b+4c+2d+e

P(2)=32+8b+4c+2d+e=34

giving

8b+4c+2d+e=2---(1)

similarly

P(-2)=32-8b+4c-2d+e=34

-8b+4c-2d+e=2--(2)

adding (1)&(2)

8c+2e=4

=> color(Blue)(4c+e=2---(3))

p(1)=2+b+c+d+e=10

giving

b+c+d+e=8--(4)

p(-1)=2-b+c-d+e=10#

giving

-b+c-d+e=8--(5)

Adding (4)&(5)

2c+2e=16

color(blue)(c+e=8--(6))

so we have

color(Blue)(4c+e=2---(3))

color(blue)(c+e=8--(6))

subtract

3c=-6

=>c=-2

:.e=8-c=8- -2=10

:.P(0)=10

Jun 11, 2017

p(0) = 10

Explanation:

From the values given, we can tell that p(x) is an even function, hence of the form:

p(x) = 2x^4+ax^2+b

Then using the given values, we have:

10 = p(1) = 2+a+b" " and hence " "a+b=8

34 = p(2) = 32+4a+b" " and hence " "4a+b=2

Subtracting the first of these equations from the second, we get:

3a=-6" " and hence " "a=-2

Then from the first equation:

b = 8-a = 8-(-2) = 10

So:

p(0) = 0+0+b = 10