Question

Q is a degree 5 polynomial that passes through the origin, has zeros i and 7 - i, and q(-1)= -520. find the equation for q​

Answer 1

Using the factor theorem, the equation for q is:

$q(x) = 4(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

The Factor Theorem states that a polynomial function with roots $x_1, x_2, ..., x_n$ is given by:

$f(x) = a(x - x_1)(x - x_2)...(x - x_n)$

In which a is the leading coefficient.

In this problem:

• Passes through the origin, so x = 0 is a zero, that is $x_1 = 0$.
• x = i is a zero, and thus, it's conjugate also is, that is, x = -i, hence $x_2 = i, x_3 = -i$

• x = 7 - i is a zero, and thus, it's conjugate also is, that is, x = 7 + i, hence $x_4 = 7 - i, x_3 = 7 + i$

Then, the equation is:

$q(x) = a(x - 0)(x - i)(x + i)(x - 7 + i)(x - 7 - i)$

$q(x) = ax(x^2 - i^2)((x-7)^2 - i^2)$

Considering that $i^2 = -1$

$q(x) = ax(x^2 + 1)(x^2 - 14x + 50)$

$q(x) = ax(x^4 - 14x^3 + 51x^2 - 14x + 50)$

$q(x) = a(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

q(-1) = -520 means that when $x = -1, q = -520$, and this is used to find a.

$q(x) = a(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

$-520 = a(-1 - 14 - 51 - 14 - 50)$

$130a = 520$

$a = \frac{520}{130}$

$a = 4$

Hence, the equation is:

$q(x) = 4(x^5 - 14x^4 + 51x^3 - 14x^2 + 50x)$

A similar problem is given at brainly.com/question/24380382