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x^(3)+7x^(2)-36
The polynomial has zeros at -6 and 2 . If the remaining zero is
z, then what is the value of
-z ?

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$x^{3}+7x^{2}-36$ The polynomial has zeros at $-6$ and $2$ . If the remaining zero is $z$ , then what is the value of $-z$ ? $\newline$ $\square$

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Find derivatives of polynomials

Full solution

Q.

x^{3}+7x^{2}-36

The polynomial has zeros at

-6

and

2

. If the remaining zero is

z

, then what is the value of

-z

?

\newline

\square

Identify Factors: Since the polynomial has zeros at $-6$ and $2$ , we can write two factors of the polynomial as $(x + 6)$ and $(x - 2)$ .
Find Third Factor: To find the third factor, we can divide the polynomial by the product of these two factors.
Perform Division: Let's perform the division: $(x^3 + 7x^2 - 36) \div ((x + 6)(x - 2))$ .
Expand Factors: First, we expand $(x + 6)(x - 2)$ to get $x^2 - 2x + 6x - 12$ , which simplifies to $x^2 + 4x - 12$ .
Use Polynomial Division: Now, we divide $x^3 + 7x^2 - 36$ by $x^2 + 4x - 12$ using polynomial long division.
Determine Quotient: The quotient will be $x + 3$ , which means the third factor of the polynomial is $(x + 3)$ .
Find Remaining Zero: Therefore, the remaining zero $z$ is $-3$ .
Calculate Negative Zero: To find the value of $-z$ , we simply negate $z$ , so $-z = -(-3) = 3$ .

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