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# Question from DragonLord

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Question from DragonLord

1.
When a polynomial $$p(x)$$ is divided by $$x+1$$  the remainder is $$5$$.
When $$p(x)$$ is divided by $$x+5$$  the remainder is $$-7$$.
Find the remainder when $$p(x)$$ is divided by $$(x+1)(x+5)$$.

2.
A polynomial of degree  $$13$$ is divided by $$d(x)$$ to give a quotient of degree 7  and a remainder of $$3x^3+4x^2-x+12$$.
What is $$deg(d)$$?

$$\begin{array}{|rcll|} \hline p(x) &=& q(x)(x+1) + 5 \quad | \quad x=-1 \\ p(-1) &=& q(x)*0 + 5 \\ \mathbf{p(-1)} &=& \mathbf{5} \\\\ p(x) &=& q(x)(x+5) -7 \quad | \quad x=-5 \\ p(-5) &=& q(x)*0 -7 \\ \mathbf{p(-5)} &=& \mathbf{-7} \\\\ p(x) &=& q(x)(x+1)(x+5) + ax+b \\ && \boxed{ (x+1)(x+5)=x^2+6x+5 \text{ is deg }2, \\ \text{so the remainder } ax+b \text{ is deg }1 } \\ p(x) &=& q(x)(x+1)(x+5) + ax+b \quad | \quad x=-1 \\ p(-1)=5 &=& 0 + -a+b \\ \mathbf{-a+b} &=& \mathbf{5} \quad \text{ or } \quad b=5+a \\ p(x) &=& q(x)(x+1)(x+5) + ax+b \quad | \quad x=-5 \\ p(-5)=-7 &=& 0 + -5a+b \\ \mathbf{-5a+b} &=& \mathbf{-7} \\\\ \mathbf{-a+b} &=& \mathbf{5} \qquad (1) \\ \mathbf{-5a+b} &=& \mathbf{-7} \qquad (2) \\ (1)-(2) : \quad 4a &=& 12 \\ \mathbf{a} &=& \mathbf{3} \\\\ b &=& 5+a \\ b &=& 5+3 \\ \mathbf{b} &=& \mathbf{8} \\ \hline \end{array}$$

The remainder when $$p(x)$$ is divided by $$(x+1)(x+5)$$ is $$\mathbf{3x+8}$$

$$\begin{array}{|rcll|} \hline \underbrace{\underbrace{p(x)}_{x^{13}+\ldots} = \underbrace{q(x)}_{x^{7}+ \ldots}*\underbrace{d(x)}_{x^6\ldots}}_{ x^{13}=x^{7}*x^{6}=x^{7+6}=x^{13}} + 3x^3+4x^2-x+12 \\ \hline \end{array}$$

$$\mathbf{deg(d) = 6}$$

Feb 5, 2020

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Thanks, heureka,  for that answer to  (1).....I didn't know how to approach that problem.....!!!

Feb 5, 2020