Question

Consider the polynomial p(x)=x^3-9x^2+18x-25, which can be rewritten as p(x)=(x-7)(x^2-2x+4)+3 . The number _[blank 1]_ is the remainder whenp(x) is divided by x-7, and so _[blank 2]_ a factor of p(x)
What is blank 1 and 2?

options:
a)7
b)is
c)is not
d)0
e)3

Answer 1

Answer:

Blank 1: 3 is the remainder

Blank 2: not a factor

Step-by-step explanation:

If p(x)=(x-7)(x^2-2x+4)+3, then dividing both sides by (x-7) gives:

$\frac{p(x)}{x-7}=(x^2-2x+4)+\frac{3}{x-7}$.

The quotient is $(x^2-2x+4)$.

The remainder is $3$.

The divisor is $(x-7)$.

The dividend is $p(x)=x^3-9x^2+18x-25$.

It is just like with regular numbers.

$\frac{11}{3}$ as a whole number is $3\frac{2}{3}$.

$3\frac{2}{3}=3+\frac{2}{3}$ where 3 is the quotient and 2 is the remainder when 11 is divided by 3.

 Here is the division just for reminding purposes:

                        3 <--quotient

                      ----

divisor->  3  |   11  <--dividend

                       -9

                        ---

                         2   <---remainder

Anyways just for fun, I would like to verify the given equation of

p(x)=(x-7)(x^2-2x+4)+3.

I would like to do by dividing myself.

I could use long division, but I have a choice to use synthetic division since we are dividing by a linear factor.

Since we are dividing by x-7, 7 goes on the outside:

         x^3-9x^2+18x  -25

7   |    1     -9        18    -25

    |            7       -14     28

     -------------------------------

          1      -2       4         3

We have confirmed what they wrote is totally correct.

The quotient is $x^2-2x+4$ while the remainder is 3.

If p/(x-7) gave a remainder of 0 then we would have said (x-7) was a factor of p.

It didn't so it isn't.

Just like with regular numbers. Is 3 a factor of 6? Yes, because the remainder of dividing 6 by 3 is 0.