Solve the equation $x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0$ . If two roots being $1$ and $7$ , Find the sum of the square of other two roots.

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Solution

If $1$ and $7$ are the roots of the given equation.

$(x - 1) (x - 7)$ are the factors of $x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0$ .

Divide $x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105$ by $x^{2} - 8 x + 7$ and find the quotient
$\begin{matrix} x^{2} + 8 x + 15 x^{2} - 8 x + 7 & x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105_x^{4}_+ - 8 x^{3}_- + 7 x^{2} - 8 x^{3} + 79 x^{2} - 176 x + 105_{+} - 8 x^{3}_{-} + 64 x^{2}_{+} - 56 x 15 x^{2} - 120 x + 105_- 15 x^{2} +_+ 120 x_{-} + 105 0 \end{matrix}$ Quotient is $x^{2} - 8 x + 15 = 0, (x - 3) (x - 5) = 0, x = 3, 5$

Sum of the square of the square of other two roots $= 3^{2} + 5^{2} = 9 + 25 = 34$ .

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Solve the equation $x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0$ . If two roots being 1 and 7, Find the sum of the square of other two roots.

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x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0

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1

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x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0

Write bi-quadratic polynomial $x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105$ as a product of two quadratic polynomial If one quadratic polynomial has roots 1 and 7.

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Solve the equation x4−16x3+86x2−176x+105=0. If two roots being 1 and 7, Find the sum of the square of other two roots. __

Solve the equation $x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0$ . If two roots being $1$ and $7$ , Find the sum of the square of other two roots.

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