The easiest terms to check are the first (8x)(2x²) = 16x³ and the last (-5)(-6) = 30. This check eliminates the first choice. The remaining choices differ only in the sign and coefficient of the squared term, so that is the one we need to find.
The squared term will be the sum of the products of factors whose degrees total 2:
(8x)(-5x) + (-5)(2x²) = -40x² -10x² = -50x²
The appropriate choice is
16x³ -50x² -23x +30
Answer:
24 red marbles
Step-by-step explanation:
Ok, why not let R = number of red marbles.
The first sentence says:
3 * R = (33 and 1/3) % * R + 64
or
3R = 100/3 %*R + 64
3R = (1/3)*R + 64
9R = R + 192
8R = 192
R = 24 red marbles
Is that dash supposed to be a minus or no
Starting off with the polynomial in standard form would be extremely difficult, but we can construct one fairly easily with the zeroes we've been given.
We know from the given zeroes that our function has the value 0 when x = 1, x = -2, and x = 2. Manipulating each equation, we can rewrite them as x - 1 = 0, x + 2 = 0, and x - 2 = 0. To construct our polynomial, we simply use all three of the expressions on the left side of the equation as factors and multiply them together, obtaining:

Notice that we can easily obtain each our three zeroes by dividing both sides by the two other factors. From here, we just need to expand the left-hand side of the equation. I'll show the work required here:
=0\\ (x^2-x+2x-2)(x-2)=0\\ (x^2+x-2)(x-2)=0\\ (x^2+x-2)x-(x^2+x-2)2=0\\ x^3+x^2-2x-(2x^2+2x-4)=0\\ x^3+x^2-2x-2x^2-2x+4=0\\ x^3+(x^2-2x^2)+(-2x-2x)+4=0\\ x^3-x^2-4x+4=0\\](https://tex.z-dn.net/?f=%28x-1%29%28x%2B2%29%28x-2%29%3D0%5C%5C%0A%5Cbig%5B%28x-1%29x%2B%28x-1%292%5Cbig%5D%28x-2%29%3D0%5C%5C%0A%28x%5E2-x%2B2x-2%29%28x-2%29%3D0%5C%5C%0A%28x%5E2%2Bx-2%29%28x-2%29%3D0%5C%5C%0A%28x%5E2%2Bx-2%29x-%28x%5E2%2Bx-2%292%3D0%5C%5C%0Ax%5E3%2Bx%5E2-2x-%282x%5E2%2B2x-4%29%3D0%5C%5C%0Ax%5E3%2Bx%5E2-2x-2x%5E2-2x%2B4%3D0%5C%5C%0Ax%5E3%2B%28x%5E2-2x%5E2%29%2B%28-2x-2x%29%2B4%3D0%5C%5C%0Ax%5E3-x%5E2-4x%2B4%3D0%5C%5C)
So, in standard form, our cubic polynomial would be