Question

H e l p ! ! ! ! :c . . . . . . . . .

Answer 1

Your answer would Be B. !!!

Explanation:

Well First you would start on the right and multiply to the left and add the x's with the numbers provided to them.

Answer 2

The answer is letter C.

Explanation:

First we have to multiplied the first term of the equaion below with the all the term of the equation above (Remember, when multiplies two terms with different exponents and coefficients, the exponents are summed and the coeffcients multiplied):

[tex] (3x^2+5x+1)*x^2 = 3x^4+5x^3+x^2 [tex/]

Then we multiply the second term of the equation below with all the terms in the equation above:

[tex] (3x^2+5x+1)*(-2x^2) = -6x^3-10x^2-2x [tex/]

We repeat the same procedure with thelast term of the equation below:

[tex] (3x^2+5x+1)*4 = 12x^2+20x+4 [tex/]

Now we have sum the terms of tha same order. For the fourth order we only have:

[tex] 3x^4 [tex/]

For the third order we have:

[tex] 5x^3 and -6x^3 = - x^3 [tex/]

For the second order

[tex] x^2+10x^2+12x^2 = -3x^2 [tex/]

For the first order:

[tex] -2x+20x = 18x [tex/]

And for the zero order:

4

Finally we sum all this terms:

[tex] = 3x^4-x^3-3x^2+18x+4