Respuesta :
Answer:
The answer is no because [tex](1)^{100}+4(1)^{99}+3 \neq 0[/tex].
Step-by-step explanation:
If [tex](x-1)[/tex] is a factor of [tex]x^{100}+4x^{99}+3[/tex], then [tex](1)^{100}+4(1)^{99}+3 \text{ will be } 0[/tex].
Let's check.
[tex](1)^{100}+4(1)^{99}+3[/tex]
[tex]1+4(1)+3[/tex]
[tex]1+4+3[/tex]
[tex]5+3[/tex]
[tex]8[/tex]
Since [tex]8 \neq 0[/tex], then [tex](x-1)[/tex] is not a factor of [tex]x^{100}+4x^{99}+3[/tex].
======================================================
So anyways more on how this works:
If [tex]\frac{P(x)}{x-c}=Q(x)+\frac{R}{x-c}[/tex] where [tex]R[/tex] is the remainder and [tex]Q(x)[/tex] is the quotient of the division, then multiplying both sides by [tex](x-c)[/tex] gives [tex]P(x)=Q(x)(x-c)+R[/tex].
We see that [tex]x-c[/tex] is a factor if the remainder is 0. (Just like we know 2 is a factor of 6 because when we divide 6 by 2 we get a remainder of 0.)
Anyways if we evaluate [tex]P(x)[/tex] for [tex]x=c[/tex] we get:
[tex]P(c)=Q(x)(c-c)+R[/tex]
[tex]P(c)=Q(x)(0)+R[/tex]
[tex]P(c)=R[/tex]
So evaluating [tex]P[/tex] for [tex]x=c[/tex] will tell us the remainder of the quotient [tex]\frac{P(x)}{x-c}[/tex] without actually performing division.
The remainder will tell us if [tex]x-c[/tex] is a factor of [tex]P[/tex]
If the remainder is 0, then the answer is yes.
If the remainder is not 0, then the answer is no.
(The question to be answer "Is this a factor of this?")
