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

Blank 1: 3 is the remainder

Blank 2: not a factor

Explanation:

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

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

`(11)/(3)` as a whole number is
`3(2)/(3)`.

`3(2)/(3)=3+(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.