If f x-1-x 171-x 191-x 231-x 231-x 291-x 341-x 411-x 431-x 47-A-B x-C x 2-then A-

Let nbsp;fx=1+x171+x191+x231+x231+x291+x341+x411+x431+x47 fx=1+x171+x191+x231+x231+x291+x341+x411+x431+x47Taking #160;out #160;1+x17, #160;1+x23 #1 ...

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Byju's Answer

Standard X

Mathematics

Fundamental Theorem of Arithmetic

If f x=1+x 17...

Question

If $f (x) = |\begin{matrix} {(1 + x)}^{17} & {(1 + x)}^{19} & {(1 + x)}^{23} \\ {(1 + x)}^{23} & {(1 + x)}^{29} & {(1 + x)}^{34} \\ {(1 + x)}^{41} & {(1 + x)}^{43} & {(1 + x)}^{47} \end{matrix}| = A + B x + C x^{2} +$ ______, then A = _.

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Solution

Let $f (x) = |\begin{matrix} {(1 + x)}^{17} & {(1 + x)}^{19} & {(1 + x)}^{23} \\ {(1 + x)}^{23} & {(1 + x)}^{29} & {(1 + x)}^{34} \\ {(1 + x)}^{41} & {(1 + x)}^{43} & {(1 + x)}^{47} \end{matrix}|$

$f (x) = |\begin{matrix} {(1 + x)}^{17} & {(1 + x)}^{19} & {(1 + x)}^{23} \\ {(1 + x)}^{23} & {(1 + x)}^{29} & {(1 + x)}^{34} \\ {(1 + x)}^{41} & {(1 + x)}^{43} & {(1 + x)}^{47} \end{matrix}| Taking out {(1 + x)}^{17}, {(1 + x)}^{23} and {(1 + x)}^{41} common from R_{1}, R_{2} and R_{3}, respectively = {(1 + x)}^{17} {(1 + x)}^{23} {(1 + x)}^{41} |\begin{matrix} 1 & {(1 + x)}^{2} & {(1 + x)}^{6} \\ 1 & {(1 + x)}^{6} & {(1 + x)}^{11} \\ 1 & {(1 + x)}^{2} & {(1 + x)}^{6} \end{matrix}| = {(1 + x)}^{81} |\begin{matrix} 1 & {(1 + x)}^{2} & {(1 + x)}^{6} \\ 1 & {(1 + x)}^{6} & {(1 + x)}^{11} \\ 1 & {(1 + x)}^{2} & {(1 + x)}^{6} \end{matrix}| Taking out {(1 + x)}^{2} and {(1 + x)}^{6} common from C_{2} and C_{3}, respectively = {(1 + x)}^{81} {(1 + x)}^{2} {(1 + x)}^{6} |\begin{matrix} 1 & 1 & 1 \\ 1 & {(1 + x)}^{4} & {(1 + x)}^{5} \\ 1 & 1 & 1 \end{matrix}| = {(1 + x)}^{89} |\begin{matrix} 1 & 1 & 1 \\ 1 & {(1 + x)}^{4} & {(1 + x)}^{5} \\ 1 & 1 & 1 \end{matrix}| = {(1 + x)}^{89} (0) (∵ if two row s are identical then the value of determinant is zero) = 0$

Hence, A = 0.

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If fx=1+x171+x191+x231+x231+x291+x341+x411+x431+x47=A+Bx+Cx2+ ______________, then A = _________.

If $f (x) = |\begin{matrix} {(1 + x)}^{17} & {(1 + x)}^{19} & {(1 + x)}^{23} \\ {(1 + x)}^{23} & {(1 + x)}^{29} & {(1 + x)}^{34} \\ {(1 + x)}^{41} & {(1 + x)}^{43} & {(1 + x)}^{47} \end{matrix}| = A + B x + C x^{2} +$ ______, then A = _.