Question

Is (x-1) a factor of x^100+4x^99+3?Can you please show steps

Answer 1

Answer:

The answer is no because $(1)^{100}+4(1)^{99}+3 \neq 0$.

Step-by-step explanation:

If $(x-1)$ is a factor of $x^{100}+4x^{99}+3$, then $(1)^{100}+4(1)^{99}+3 \text{ will be } 0$.

Let's check.

$(1)^{100}+4(1)^{99}+3$

$1+4(1)+3$

$1+4+3$

$5+3$

$8$

Since $8 \neq 0$, then $(x-1)$ is not a factor of $x^{100}+4x^{99}+3$.

======================================================

So anyways more on how this works:

If $\frac{P(x)}{x-c}=Q(x)+\frac{R}{x-c}$ where $R$ is the remainder and $Q(x)$ is the quotient of the division, then multiplying both sides by $(x-c)$ gives $P(x)=Q(x)(x-c)+R$.

We see that $x-c$ 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 $P(x)$ for $x=c$ we get:

$P(c)=Q(x)(c-c)+R$

$P(c)=Q(x)(0)+R$

$P(c)=R$

So evaluating $P$ for $x=c$ will tell us the remainder of the quotient $\frac{P(x)}{x-c}$ without actually performing division.

The remainder will tell us if $x-c$ is a factor of $P$

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?")