The $$expression14x+11+4x+127$$−$$1$$−$$4x+127is$$ a polynomial in x of degree.

We have,$$14x+11+4x+127$$−$$1$$−$$4x+127=1274x+12$$ $$7C14x+1+7C3(4x+1)3+7C5(4x+1)5+7C7(4x+1)7$$ $$=126$$ $$7C1+7C3(4x+1)+7C5(4x+1)2+7C7(4x+1)3Clearly$$, it is a polynomial of degree $$3$$.

Binomial theorem for positive integral Index

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Question

The expression
$\frac{1}{\sqrt{4 x + 1}} \{{(1 + \frac{\sqrt{4 x + 1}}{2})}^{7} - {(1 - \frac{\sqrt{4 x + 1}}{2})}^{7}\}$
is a polynomial in $x$ of degree.

Moderate

Solution

We have,
$\begin{array}{l} \frac{1}{\sqrt{4 x + 1}} \{{(1 + \frac{\sqrt{4 x + 1}}{2})}^{7} - {(1 - \frac{\sqrt{4 x + 1}}{2})}^{7}\} \\ = \frac{1}{2^{7} \sqrt{4 x + 1}} [2 \{^{7} C_{1} \sqrt{4 x + 1} +^{7} C_{3} (\sqrt{4 x + 1})^{3} +^{7} C_{5} (\sqrt{4} x + 1)^{5} +^{7} C_{7} (\sqrt{4 x + 1})^{7}\}] \end{array}$
$= \frac{1}{2^{6}} \{^{7} C_{1} +^{7} C_{3} (4 x + 1) +^{7} C_{5} (4 x + 1)^{2} +^{7} C_{7} (4 x + 1)^{3}\}$
Clearly, it is a polynomial of degree 3.

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

The expression14x+11+4x+127−1−4x+127is a polynomial in x of degree.

We have,14x+11+4x+127−1−4x+127=1274x+12 7C14x+1+7C3(4x+1)3+7C5(4x+1)5+7C7(4x+1)7 =126 7C1+7C3(4x+1)+7C5(4x+1)2+7C7(4x+1)3Clearly, it is a polynomial of degree 3.

Ready to Test Your Skills?

free study material

The expression
$\frac{1}{\sqrt{4 x + 1}} \{{(1 + \frac{\sqrt{4 x + 1}}{2})}^{7} - {(1 - \frac{\sqrt{4 x + 1}}{2})}^{7}\}$
is a polynomial in $x$ of degree.