Given a polynomial px-x5-9 x4-24 x3-52 x-48- The zeroes of the polynomial are -2-2 and 2 - Find the other 2 zeroes-A- -3 and -4B- 3 and -4C- 3 and 4D- -3 and 4

The correct option is C 3 and 4 Given zeroes are √(2), √(2) and 2 g(x) = (x √(2))(x + √(2))(x 2) =(x^2 2)(x 2) = x^3 2x^2 2x+4 Dividing p(x) by g(x) enclos ...

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Byju's Answer

Standard VII

Mathematics

Dividing a Polynomial by a Polynomial

Given a polyn...

Question

Given a polynomial p(x) = $x^{5} - 9 x^{4} + 24 x^{3} - 52 x + 48$ . The zeroes of the polynomial are $\sqrt{2}, - \sqrt{2}$ and 2. Find the other 2 zeroes.

-3 and 4

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-3 and -4

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3 and -4

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3 and 4

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Solution

The correct option is D
3 and 4

Given zeroes are $\sqrt{2}, - \sqrt{2}$ and 2

g(x) = $(x - \sqrt{2}) (x + \sqrt{2}) (x - 2)$ = $(x^{2} - 2) (x - 2)$ = $x^{3} - 2 x^{2} - 2 x + 4$

Dividing p(x) by g(x)

$\begin{matrix} x^{2} - 7 x + 12 x^{3} - 2 x^{2} - 2 x + 4 x^{5} - 9 x^{4} + 24 x^{3} - 6 x^{2} + 52 x + 48 x^{5} - 2 x^{4} - 2 x^{3} + 4 x^{2} - --------------------- - - 7 x^{4} + 26 x^{3} - 10 x^{2} + 52 x - 7 x^{4} + 14 x^{3} - 14 x^{2} - 28 x - --------------------------- - 12 x^{3} - 24 x^{2} + 24 x + 48 12 x^{3} - 24 x^{2} - 24 x + 48 - ------------------------ - 0 \end{matrix}$

q(x) = $x^{2} - 7 x + 12$

p(x) = g(x) h(x)

= $(x^{3} - 2 x^{2} - 2 x + 4) (x^{2} - 7 x + 12)$

The other two zeroes are in $x^{2} - 7 x + 12$

= $x^{2} - 4 x - 3 x + 12$ = $x (x - 4) - 3 (x - 4)$ = $(x - 3) (x - 4)$

Hence, the zeroes are 3 and 4

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Given a polynomial p(x) = x5−9x4+24x3−52x+48. The zeroes of the polynomial are √2,−√2 and 2. Find the other 2 zeroes.

Given a polynomial p(x) = $x^{5} - 9 x^{4} + 24 x^{3} - 52 x + 48$ . The zeroes of the polynomial are $\sqrt{2}, - \sqrt{2}$ and 2. Find the other 2 zeroes.