Polynomial Division: where does the remainder go?

Question

If

P(x) = x ⁴ + 4x ³ - 14x ² + 36x + 45

is divided by x + 5

and the remainder that I found is -1250, would it go at the end of the quotient? Like this:

x ³ - 9x ² - 59x + 259 - 1250 ?

*_{...forgot how to do this stuff :\}

Answer 1

Since $P(-5)=-360$ and since$$\frac{P(x)+360}{x+5}=x^3-x^2-9x+81,$$you have $P(x)=(x^3-x^2-9x+81)(x+5)+360$.

Answer 2

$\,P(x)=q(x).g(x)+r(x)$ :.
$(x^4+4x^3-14x^2+36x+45) = (x^3-x^2-9x+81).(x+5) + (-360)$.

I hope you asked for this, otherwise pardon.

Alternatively use remainder theorem... to find remainder only. $P(-5)=-360.$

Answer 3

I suggest you do not trust your memory, think of division from first principles. How does integer division work? If you try to divide $\frac{7}{5}$ you know that the quotient is $1$, but how do you find the remainder? You multiply $1\cdot 5=5$ but you still need $2$ to reach $7$. In other words: $$ 7 = 1 \cdot 5 + 2$$ or, in general, if you divide $\frac{N}{D}$ obtaining the quotient $Q$ and the remainder $R$, it is true than: $$ N = Q \cdot D + R$$

Now, if you divide both sides of this equation by $D$ you recover the original fraction: $$\frac{N}{D}=Q+\frac{R}{D}$$

It works the same with polynomials, so in your case: $$\frac{P(x)}{x+5}=Q(x)+\frac{R}{x+5}$$ where $R=-360$ as other answers have pointed out.

Answer 4

HINT: the right result should be $${x}^{3}-{x}^{2}-9\,x+81-360\, \left( x+5 \right) ^{-1}$$the remainder is given by $-360$