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
sjc
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#

George C.
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#

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