+0

# any help is appreciated few tough algebra 2 questions :(

0
304
2

1. Let $$f(x) = x^{10}+5x^9-8x^8+7x^7-x^6-12x^5+4x^4-8x^3+12x^2-5x-5$$.

Without using long division, find the remainder when $$f(x)$$ is divided by $$x^2-1$$.

2. Let $$f(x)$$ be a polynomial such that $$f(0) = 4,$$ $$f(1) = 5,$$ and $$f(2)=10.$$ Find the remainder when $$f(x)$$ is divided by $$x(x - 1)(x - 2).$$

Thank you very much for your help!

Aug 11, 2020

#1
+25984
+2

1.

$$f(x) = x^{10}+5x^9-8x^8+7x^7-x^6-12x^5+4x^4-8x^3+12x^2-5x-5$$

Without using long division, find the remainder $$(=r)$$  when $$f(x)$$  is divided by $$x^2-1$$.

Let $$x^2-1 = (x-1)(x+1)$$

$$\begin{array}{|lrclrcl|} \hline & f(x) &=& q(x)(x-1)(x+1) + r \quad | \quad \mathbf{r=ax+b} \\ & \mathbf{ f(x) } &=& \mathbf{ q(x)(x-1)(x+1) + ax+b } \\ \hline x=1:& \mathbf{ f(1) } &=& \mathbf{ q(1)(1-1)(1+1) + a*1+b } \\ & f(1)&=& q(1)*0 + a+b \\ & f(1)&=& a+b \\ &&& \small {f(1) = 1*1^{10}+5*1^9-8*1^8+7*1^7-1^6-12*1^5+4*1^4-8*1^3+12*1^2-5*1-5 }\\ &&& f(1) = 1+5-8+7-1-12+4-8+12-5-5 \\ &&& f(1) = -8+7+4-8-5 \\ &&& f(1) = 11-21 \\ &&& f(1) = -10 \\ & -10 &=& a+b \\ & \mathbf{a+b} &=& \mathbf{-10} \qquad (1) \\ \hline x=-1:& \mathbf{ f(-1) } &=& \mathbf{ q(-1)(-1-1)(-1+1) + a*(-1)+b } \\ & f(-1)&=& q(-1)*0 - a+b \\ & f(-1)&=& - a+b \\ &&& \small{f(-1) = 1*(-1)^{10}+5*(-1)^9-8*(-1)^8+7*(-1)^7-(-1)^6-12*(-1)^5+4*(-1)^4-8*(-1)^3+12*(-1)^2-5*(-1)-5} \\ &&& f(-1) = 1-5-8-7-1+12+4+8+12+5-5 \\ &&& f(-1) =12+4 \\ &&& f(-1) = 16 \\ & 16 &=& - a+b \\ & \mathbf{-a+b} &=& \mathbf{16} \qquad (2)\\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & \mathbf{a+b} &=& \mathbf{-10} \qquad (1) \\ & \mathbf{-a+b} &=& \mathbf{16} \qquad (2)\\ \hline (1)+(2): & a+b-a+b&=&-10+16 \\ & 2b&=& 6 \\ & \mathbf{b} &=& \mathbf{3} \\ \hline (1)-(2): & a+b-(-a+b) &=& -10-16 \\ & 2a &=& -26 \\ & \mathbf{a} &=& \mathbf{-13} \\ \hline & r &=& ax+b \\ & \mathbf{ r } &=& \mathbf{-13x+3} \\ \hline \end{array}$$

The remainder when $$f(x)$$ is divided by $$x^2-1$$ is $$\mathbf{-13x+3}$$

Aug 12, 2020
edited by heureka  Aug 12, 2020
#2
+25984
+2

2.
Let $$f(x)$$  be a polynomial such that $$f(0) = 4$$, $$f(1) = 5$$, and $$f(2)=10$$.
Find the remainder $$(=r)$$ when $$f(x)$$ is divided by $$x(x - 1)(x - 2)$$.

$$\begin{array}{|lrclrcl|} \hline & f(x) &=& q(x)*x(x - 1)(x - 2) + r \quad | \quad \mathbf{r=ax^2+bx+c } \\ & \mathbf{ f(x) } &=& \mathbf{ q(x)*x(x - 1)(x - 2) + ax^2+bx+c } \\ \hline x=0:& \mathbf{ f(0) } &=& \mathbf{ q(0)*0(0 - 1)(0 - 2) + a*0^2+b*0+c } \\ & f(0)&=& q(0)*0 +c \\ & f(0)&=& c \quad | \quad f(0) = 4 \\ & 4&=& c \\ & \mathbf{c} &=& \mathbf{4} \\ \hline x=1:& \mathbf{ f(1) } &=& \mathbf{ q(1)*1(1 - 1)(1 - 2) + a*1^2+b*1+c } \\ & f(1)&=& q(1)*0 a+b+4 \\ & f(1)&=& a+b+4 \quad | \quad f(1) = 5 \\ & 5&=& a+b+4 \\ & \mathbf{a+b} &=& \mathbf{1} \qquad (1) \\ \hline x=2:& \mathbf{ f(2) } &=& \mathbf{ q(2)*2(2 - 1)(2 - 2) + a*2^2+b*2+c } \\ & f(2)&=& q(2)*0 +4a+2b+4 \\ & f(2)&=& 4a+2b+4 \quad | \quad f(2) = 10 \\ & 10&=& 4a+2b+4 \quad | \quad :2 \\ & 5&=& 2a+b+2 \\ & \mathbf{2a+b} &=& \mathbf{3} \qquad (2) \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & \mathbf{a+b} &=& \mathbf{1} \qquad (1) \\ & \mathbf{2a+b} &=& 3 \qquad (2) \\ \hline (2)-(1): & 2a+b -(a+b) &=& 3-1 \\ & \mathbf{a} &=& \mathbf{2} \\ \hline & b &=& 1-a \\ & b &=& 1-2 \\ & \mathbf{b} &=& \mathbf{-1} \\ \hline & r &=& ax^2+bx+c \\ & \mathbf{ r } &=& \mathbf{2x^2-x+4} \\ \hline \end{array}$$

The remainder when $$f(x)$$ is divided by $$x(x - 1)(x - 2)$$ is  $$\mathbf{2x^2-x+4}$$

Aug 12, 2020