Question about using the Remainder Theorem for Quadratic Divisors

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I have the following Polynomials:

$P(x)$ = $5x^4$ + $5x^3$ - $3x^2$ - $3x$ - $2$

$Q_{1}(x)$ = $x - 4$

$Q_{2}(x)$ = $2x^2$ + $5x$ + $3$

I have to let $D_{1}(x)$, $D_{2}(x)$ and $R_{2}(x)$ be polynomials such that:

$P(x)$ = $Q_{1}(x)$ * $D_{1}(x)$ + $R_{1}(x)$

$P(x)$ = $Q_{2}(x)$ * $D_{2}(x)$ + $R_{2}(x)$

I have to use the long division algorithm to express $P(x)$ in the form:

$Q_{2}(x)$ * $D_{2}(x)$ + $R_{2}(x)$

I used the Remainder Theorem to find $R_{2}(x)$ first.

Since, the divisor $Q_{2}(x)$ = $2x^2$ + $5x$ + $3$.

I've factorised and solved the quadratic to get:

$(2x + 3)(x + 1) = 0$

Where $x = -1.5$ or $x = -1$

How would I know which of these to use in getting the correct Remainder, $R_{2}(x)$?

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You don't use the Remainder Theorem. The remainder is going to be a linear polynomial, not a number. You do like it says: you use long division, you divide $P$ by $Q_2$, getting a quotient and a remainder. Try it!