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

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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.

How do you determine p(c) given p(x)=x^4+x^3-6x^2-7x-7p(x)=x^4+x^3-6x^2-7x-7 and c=-sqrt7c=-sqrt7?