How do you determine p(c) given #p(x)=x^4+x^3-6x^2-7x-7# and #c=-sqrt7#?

Question

1945 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-06T10:33:23

#-14#

Grab your calculator... Because this is a REAL doozy....

Since:
#c=-sqrt7#
And:
#p(x) =x^4+x^3-6x^2-7x-7#

Substitute
#c=-sqrt7#
Into # p(x) #, which is:
#p(x)= (-sqrt7)^4+(-sqrt7)^3-6(-sqrt7)^2 - 7(-sqrt7)-7 #

Simplifying gives

#(-7)^2-(7)^(3/2)-6(7)-7(7)^(1/2) - 7#

Using a calculator,
#7^(3/2)=18.520#

And
#7^(1/2) = 2.646#

Solving all the stuffs
#49-18.520-42+7(2.646)-7#

#49-42-7#

#=0#

Yay! DONE!

Answer 2 · 2017-07-06T10:36:11

#0#

What we can do is plug in the value #-sqrt7# in for each #x#:

#p(c) = (-sqrt7)^4 + (-sqrt7)^3 - 6(-sqrt7)^2 - 7(-sqrt7) - 7#

#= 49 - 7sqrt7 - 42 + 7sqrt7 - 7#

#= 49 - 42 - 7 = color(blue)(0#

Thus, the value "#-sqrt7#" is a zero of this polynomial function.