Respuesta :
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:
[tex]\frac{p(x)}{x-7}=(x^2-2x+4)+\frac{3}{x-7}[/tex].
The quotient is [tex](x^2-2x+4)[/tex].
The remainder is [tex]3[/tex].
The divisor is [tex](x-7)[/tex].
The dividend is [tex]p(x)=x^3-9x^2+18x-25[/tex].
It is just like with regular numbers.
[tex]\frac{11}{3}[/tex] as a whole number is [tex]3\frac{2}{3}[/tex].
[tex]3\frac{2}{3}=3+\frac{2}{3}[/tex] 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 [tex]x^2-2x+4[/tex] 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.
