Use the remainder theorem to determine if x = -4 is a zero of the following polynomial, and find the quotient and the remainder. p(x) = x^2 - 169 a. yes, x = -4 is a zero of the polynomial. the quotient is x + 4, and the remainder is 0. b. no, x = -4 is not a zero of the polynomial. the quotient is x + 4 and the remainder is -153. c. yes, x = -4 is a zero of the polynomial. the quotient is x - 4, and the remainder is 153. d. no, x = -4 is not a zero of the polynomial. the quotient is x - 4 and the remainder is -153.

choice D

x = -4 is not a zero of the polynomial

quotient = x-4

remainder = -153

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Work Shown:

q = quotient = ax+b

r = remainder

r = p(-4), aka the function output of p(x) when the value x = -4 is plugged in.

p(x) = x^2 - 169

p(-4) = (-4)^2 - 169

p(-4) = 16 - 169

p(-4) = -153 is the remainder, so r = -153.

Since the remainder is nonzero, this means x = -4 is not a root or zero or x intercept of p(x). This also means x+4 is not a factor of p(x).

p(x)/(x+4) = q + r/(x+4)

p(x) = q(x+4) + r

x^2 - 169 = q(x+4) - 153

x^2 - 169 = (ax+b)(x+4) - 153 ... replace 'q' with ax+b

x^2 + 0x - 169 = ax^2 + 4ax + bx + 4b - 153

x^2 + 0x - 169 = ax^2 + (4a+b)x + (4b-153)

We see that x^2 on the left side matches with ax^2 on the right side, so a = 1.

The middle terms 0x and (4a+b)x match up, so 0 = 4a+b. Use a = 1 to find that b = -4. So the quotient is q = ax+b = 1x+(-4) = x-4

You can also use polynomial long division to achieve the same goal. See the attached image below.

step-by-step explanation:

i believe its "c" because all of the other options include a but it could be d because there is a way of financing with cash at certain places but its uncommon.

so then what are you asking?

step-by-step explanation: