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creativ13 [48]
3 years ago
7

X-3÷4x^4-8x^3-18x^2+17x+3​

Mathematics
1 answer:
givi [52]3 years ago
8 0

<u>Answer:</u>

The simplication of x-3÷4x^4-8x^3-18x^2+17x+3 is \frac{1}{\left(4 x^{3}+4 x^{2}-6 x-1\right)}

<u>Solution:</u>

\text { Given, expression is }(x-3) \div\left(4 x^{4}-8 x^{3}-18 x^{2}+17 x+3\right)

Now we have to simplify the given expression,

For that, we have to factorize the denominator.

\text { So, } 4 x^{4}-8 x^{3}-18 x^{2}+17 x+3

\Rightarrow 4 x^{4}+\left(-12 x^{3}+4 x^{3}\right)+\left(-12 x^{2}-6 x^{2}\right)+(18 x-x)+3

By grouping terms we get,

\rightarrow\left(4 x^{4}-12 x^{3}\right)+\left(4 x^{3}-12 x^{2}\right)-\left(6 x^{2}-18 x\right)-(x-3)

By taking the common terms,

\begin{array}{l}{\rightarrow 4 x^{3}(x-3)+4 x^{2}(x-3)-6 x(x-3)-(x-3)} \\\\ {\rightarrow(x-3)\left(4 x^{3}+4 x^{2}-6 x-1\right)}\end{array}

\begin{array}{l}{\text { Now, }(x-3) \div\left(4 x^{4}-8 x^{3}-18 x^{2}+17 x+3\right) \rightarrow \frac{x-3}{\left(4 x^{4}-8 x^{3}-18 x^{2}+17 x+3\right)}} \\\\ {\quad \rightarrow \frac{x-3}{(x-3)\left(4 x^{3}+4 x^{2}-6 x-1\right)}} \\\\ {\quad \rightarrow \frac{1}{\left(4 x^{3}+4 x^{2}-6 x-1\right)}}\end{array}

Hence, the simplified expression is  \frac{1}{\left(4 x^{3}+4 x^{2}-6 x-1\right)}

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  appropriately writing the proportion can reduce or eliminate steps required to solve it

Step-by-step explanation:

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This equation can be written as proportions in 3 other ways:

  \dfrac{B}{A}=\dfrac{D}{C}\qquad\dfrac{A}{C}=\dfrac{B}{D}\qquad\dfrac{C}{A}=\dfrac{D}{B}

  Effectively, the proportion can be written upside-down and sideways, as long as the corresponding parts are kept in the same order.

I find this most useful to ...

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Usual method:

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Multiplying by the denominator:

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<u>Example 2</u>:

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The point here is that if you have a choice when you write the initial proportion, you can make the choice to write it with x in the numerator and 1 in the denominator.

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