Question

1. the cubic $x^{3}+ax^{2}-3x + 2b$ has a remainder of 4 when divided by $(x - 3)$, and $x^{3}+ax^{2}-x - 3b$ has a remainder of -8 when divided by $(x + 1)$. determine the value of $a$ and $b$.

m). use binomial theorem to find the constant term in the expansion of $(2x-\frac{1}{x^{2}})^{9}$.
n). use binomial theorem to expand and simplify $(3x - 2y)^{4}$.

Explanation:

Step1: Recall the binomial theorem

The binomial theorem for \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\).

Step2: For part m)

For \((2x-\frac{1}{x^2})^9\), we have \(a = 2x\), \(b=-\frac{1}{x^2}\) and \(n = 9\). The general - term \(T_{r + 1}=\binom{9}{r}(2x)^{9 - r}(-\frac{1}{x^2})^{r}=\binom{9}{r}(2)^{9 - r}(x)^{9 - r}(-1)^{r}(x^{-2r})=\binom{9}{r}(2)^{9 - r}(-1)^{r}x^{9 - r-2r}=\binom{9}{r}(2)^{9 - r}(-1)^{r}x^{9 - 3r}\). To find the constant term, set the power of \(x\) equal to 0, i.e., \(9-3r = 0\), so \(r = 3\). Then \(T_{4}=\binom{9}{3}(2)^{9 - 3}(-1)^{3}\). Calculate \(\binom{9}{3}=\frac{9!}{3!(9 - 3)!}=\frac{9\times8\times7}{3\times2\times1}=84\), \((2)^{6}=64\). So \(T_{4}=84\times64\times(-1)=- 5376\).

Step3: For part n)

For \((3x-2y)^4\), using the binomial theorem \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\) with \(a = 3x\), \(b=-2y\) and \(n = 4\). \(T_{r + 1}=\binom{4}{r}(3x)^{4 - r}(-2y)^{r}=\binom{4}{r}(3)^{4 - r}(x)^{4 - r}(-2)^{r}(y)^{r}\).
\((3x-2y)^4=\binom{4}{0}(3x)^{4}(-2y)^{0}+\binom{4}{1}(3x)^{3}(-2y)^{1}+\binom{4}{2}(3x)^{2}(-2y)^{2}+\binom{4}{3}(3x)^{1}(-2y)^{3}+\binom{4}{4}(3x)^{0}(-2y)^{4}\).
\(\binom{4}{0}=1\), \((3x)^{4}=81x^{4}\); \(\binom{4}{1}=\frac{4!}{1!(4 - 1)!}=4\), \((3x)^{3}=27x^{3}\), \((-2y)^{1}=-2y\), so \(\binom{4}{1}(3x)^{3}(-2y)^{1}=4\times27x^{3}\times(-2y)=-216x^{3}y\); \(\binom{4}{2}=\frac{4!}{2!(4 - 2)!}=6\), \((3x)^{2}=9x^{2}\), \((-2y)^{2}=4y^{2}\), so \(\binom{4}{2}(3x)^{2}(-2y)^{2}=6\times9x^{2}\times4y^{2}=216x^{2}y^{2}\); \(\binom{4}{3}=\frac{4!}{3!(4 - 3)!}=4\), \((3x)^{1}=3x\), \((-2y)^{3}=-8y^{3}\), so \(\binom{4}{3}(3x)^{1}(-2y)^{3}=4\times3x\times(-8y^{3})=-96xy^{3}\); \(\binom{4}{4}=1\), \((-2y)^{4}=16y^{4}\).
So \((3x - 2y)^4=81x^{4}-216x^{3}y + 216x^{2}y^{2}-96xy^{3}+16y^{4}\).

Answer:

m) The constant term of \((2x-\frac{1}{x^2})^9\) is \(-5376\).
n) \((3x - 2y)^4=81x^{4}-216x^{3}y + 216x^{2}y^{2}-96xy^{3}+16y^{4}\)