1. Find the quotient and remainder when \[ x^{3 n}-x^{2 n}+3 x^{n}-10 \text { is divided by } x^{n}-2 \] 2. Find number \( a \) and \( b \) so that \( x-2 \) is a factor of \[ f(x)=x^{4}-2 a x^{3}+a x^{2}-x+b \text { and } f(-1)=3 \] 3. Find the polynomial \( f(x)=a(x-2)^{2}+b(2 x+1)^{2} \) is divided by \( x+1 \) the and \( f(1)=10 \). Then find the values of \( a \) and \( b \). 4. Simplify A. \( \left(2^{-3}+4^{-1}\right)^{-3} \) B. \( \frac{a}{a^{3}} /\left(2-\frac{1}{1+\frac{1}{a-2}}\right) \) 5. Solve the following a. \( 3(243 / 32)^{2 x-2}=2\left(\frac{16}{81}\right)^{\frac{3 x}{2}-1} \) b. \( 5^{x^{2}+x}=25 \) d. \( 2^{4 x-2} \leq 1024 \) e. \( (1 / 100)^{x-5} \geq(0 \) 6. draw the graphs of \( f(x) \) and \( g(x) \) on the same coordinate plane and discuss its graphs. a. \( f(x)=(4 / 5)^{x} \) and \( g(x)=(5 / 4)^{x} \) b. \( f(x)=\left(\log 5^{x}\right. \) and \( g(x)=\log \left(\frac{1}{5}\right)^{x} \) 7. find the domain of the following logarithmic equation. a. \( f(x)=\log 5^{(3-x)}+\log 5^{(x+3)}=\log 5^{25} \) b. \( \log 5^{x^{2}-2 x} \) c. \( \log x^{-3 x+1} \) 8. for \( a>0, a \neq 1 \) if \( \log a^{2}=b \) and \( \log a^{3}=c \), then find \( \log a^{36}, \log a^{48}, \log (1 / 2 \) 9. solve for \( \mathrm{x} \)
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By using polynomial division, we find that the quotient is \(x^{2n} - x^n + 5\) and the remainder is 0. Show more…
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Determine the remainder when $\left(x^{3}-2 x^{2}-5 x+6\right)$ is divided by (a) $(x-1)$ and (b) $(x+2)$. Hence factorise the cubic expression. (a) When $\left(x^{3}-2 x^{2}-5 x+6\right)$ is divided by $(x-1)$, the remainder is given by $a p^{3}+b p^{2}+c p+d$, where $a=1, b=-2, c=-5, d=6$ and $p=1$ i.e. the remainder $=(1)(1)^{3}+(-2)(1)^{2}$ $$ \begin{gathered} +(-5)(1)+6 \\ =1-2-5+6=0 \end{gathered} $$ Hence $(x-1)$ is a factor of $\left(x^{3}-2 x^{2}-5 x+6\right)$ (b) When $\left(x^{3}-2 x^{2}-5 x+6\right)$ is divided by $(x+2)$, the remainder is given by $$ \begin{aligned} (1)(-2)^{3}+(-2)(-2)^{2} &+(-5)(-2)+6 \\ &=-8-8+10+6=0 \end{aligned} $$ Hence $(x+2)$ is also a factor of $\left(x^{3}-2 x^{2}-\right.$ $5 x+6)$. Therefore $(x-1)(x+2)(x)=x^{3}-$ $2 x^{2}-5 x+6$. To determine the third factor (shown blank) we could (i) divide $\left(x^{3}-2 x^{2}-5 x+6\right)$ by $(x-1)(x+2)$ or (ii) use the factor theorem where $f(x)=$ $x^{3}-2 x^{2}-5 x+6$ and hoping to choose a value of $x$ which makes $f(x)=0$ or (iii) use the remainder theorem, again hoping to choose a factor $(x-p)$ which makes the remainder zero.(i) Dividing $\left(x^{3}-2 x^{2}-5 x+6\right)$ by $\left(x^{2}+x-2\right)$ gives: $$ \begin{array}{r} x ^ { 2 } + x - 2 \longdiv { x ^ { 3 } - 2 x ^ { 2 } - 5 x + 6 } \\ \frac{x^{3}+x^{2}-2 x}{-3 x^{2}-3 x}+6 \\ \frac{-3 x^{2}-3 x+6}{} \end{array} $$ Thus $\left(x^{3}-2 x^{2}-5 x+6\right)$ $$ =(x-1)(x+2)(x-3) $$ (ii) Using the factor theorem, we let $$ f(x)=x^{3}-2 x^{2}-5 x+6 $$ Then $f(3)=3^{3}-2(3)^{2}-5(3)+6$ $$ =27-18-15+6=0 $$ Hence $(x-3)$ is a factor. (iii) Using the remainder theorem, when $\left(x^{3}-2 x^{2}-5 x+6\right)$ is divided by $(x-3)$, the remainder is given by $a p^{3}+b p^{2}+c p+d, \quad$ where $\quad a=1$ $b=-2, c=-5, d=6$ and $p=3$ Hence the remainder is: $$ \begin{aligned} 1(3)^{3}+(-2)(3)^{2} &+(-5)(3)+6 \\ &=27-18-15+6=0 \end{aligned} $$ Hence $(x-3)$ is a factor. Thus $\left(x^{3}-2 x^{2}-5 x+6\right)$ $$ =(x-1)(x+2)(x-3) $$
1. Which one of the graphs is that of $y=x^{3}-3 x^{2}-6 x+8 ?$ 2. Which one of the graphs is that of $y=x^{4}+7 x^{3}-5 x^{2}-75 x ?$ 3. How many real zeros does the graph in C have? 4. Which one of $\mathrm{C}$ and $\mathrm{D}$ is the graph of $y=-x^{3}+9 x^{2}-27 x+17 ?$ 5. Which of the graphs cannot be that of a cubic polynomial function? 6. Which one of the graphs is that of a function whose range is $\operatorname{not}(-\infty, \infty) ?$ 7. The function defined by $f(x)=x^{4}+7 x^{3}-5 x^{2}-75 x$ has the graph shown in $\mathbf{B}$. Use the graph to factor the polynomial. 8. The function defined by $f(x)=-x^{5}+36 x^{3}-22 x^{2}-147 x-90$ has the graph shown in D. Use the graph to factor the polynomial.
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Polynomial Functions: Graphs, Applications, and Models
Expand $\ln (1+x)$ to five terms. $$ \begin{array}{cc} f(x)=\ln (1+x) & f(0)=\ln (1+0)=0 \\ f^{\prime}(x)=\frac{1}{(1+x)} \quad f^{\prime}(0)=\frac{1}{1+0}=1 \\ f^{\prime \prime}(x)=\frac{-1}{(1+x)^{2}} & f^{\prime \prime}(0)=\frac{-1}{(1+0)^{2}}=-1 \\ f^{\prime \prime \prime}(x)=\frac{2}{(1+x)^{3}} & f^{\prime \prime \prime}(0)=\frac{2}{(1+0)^{3}}=2 \\ f^{\text {iv }}(x)=\frac{-6}{(1+x)^{4}} \quad f^{\mathrm{iv}}(0)=\frac{-6}{(1+0)^{4}}=-6 \\ f^{\mathrm{v}}(x)=\frac{24}{(1+x)^{5}} \quad f^{\mathrm{v}}(0)=\frac{24}{(1+0)^{5}}=24 \end{array} $$ Substituting these values into equation (5) gives: $$ \begin{aligned} f(x)=\ln (1+x)=& 0+x(1)+\frac{x^{2}}{2 !}(-1) \\ &+\frac{x^{3}}{3 !}(2)+\frac{x^{4}}{4 !}(-6)+\frac{x^{5}}{5 !}(24) \end{aligned} $$ 1.e. $\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots$
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