Given a polynomial that has zeros of −4, 9i, and −9i and has a value of 492 when x=−1. Write the polynomial in standard form axn+bxn−1+…. Answer using reduced fractions when necessary.

The polynomial that has zeros of −4, 9i, and −9i, and has a value 492 when x = -1 is:

2x³ + 8x² + 162x + 648 = 0

Step-by-step explanation:

Given that the polynomial that has zeros of −4, 9i, and −9i, we can say that

x = -4 => x + 4 = 0

x = 9i and -9i => x = ±9i => x² = (9i)²

Because i² = -1, x² = -81

=> x² + 81 = 0

So together, we can write the polynomial as

(x + 4)(x² + 81) = 0

Expanding the bracket, we have

x³ + 81x + 4x² + 324 = 0

x³ + 4x² + 81x + 324 = 0

Again, we are told that the polynomial is 492 at x = -1.

Put x = -1 in

x³ + 4x² + 81x + 324 = 0

(-1)³ + 4(-1)² + 81(-1) + 324

-1 + 4 - 81 + 324

= 246 ≠ 492

But 492 = 2×246

Since multiplying the polynomial by 2 doesn't change anything, our result is set.

Now,

2(x³ + 4x² + 81x + 324) = 2(0)

2x³ + 8x² + 162x + 648 = 0

When x = -1

2(-1)³ + 8(-1)² + 162(-1) + 648

-2 + 8 - 162 + 648

= 492

As required.

10

step-by-step explanation:

7+3=10

the first one, all those points on the line just make a strait line.