Question

answer me ok plz aur b question h ese solve krne ke baad aur milenge

Answer 1

Finding zeroes

Solution.

The given polynomial is

$\quad f(x)=x^{4}-6x^{3}-26x^{2}+138x-35$

Given that $(2+\sqrt{3})$ and $(2-\sqrt{3})$ are the two zeroes of $f(x)$,

then $x-(2+\sqrt{3}),\:x-(2-\sqrt{3})$ are factors of $f(x)$

i.e., $\{x-(2+\sqrt{3})\}\{x-(2-\sqrt{3})\}$ is a factor of $f(x)$

i.e., $(x^{2}-4x+1)$ is a factor of $f(x)$

$\therefore x^{4}-6x^{3}-26x^{2}+138x-35$

$=x^{4}-4x^{3}+x^{2}-2x^{3}+8x^{2}-2x-35x^{2}+140x-35$

$=x^{2}(x^{2}-4x+1)-2x(x^{2}-4x+1)-35(x^{2}-4x+1)$

$=(x^{2}-4x+1)(x^{2}-2x-35)$

So the other factor of $f(x)$ is $(x^{2}-2x-35)$

$\therefore x^{2}-2x-35$

$=x^{2}-7x+5x-35$

$=x(x-7)+5(x-7)$

$=(x-7)(x+5)$

$\Rightarrow x=7,\:-5$

$\therefore$ the other two zeroes of $f(x)$ are $7,\:(-5).$

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