Login/Sign Up
Home » Find all the zeroes of , if it is given that two of its zeroes are and
Find all the zeroes of \left(x^{4}+x^{3}-23 x^{2}-3 x+60\right), if it is given that two of its zeroes are \sqrt{3} and -\sqrt{3}
CBSE | Class 10 | Excercise 2B | Maths | Polynomials | RS Aggarwal
Find all the zeroes of \left(x^{4}+x^{3}-23 x^{2}-3 x+60\right), if it is given that two of its zeroes are \sqrt{3} and -\sqrt{3}
 Let f(x)=x4+x3–23x2–3x+60
\text { Let } f(x)=x^{4}+x^{3}-23 x^{2}-3 x+60

Since \sqrt{3} and -\sqrt{3} are the zeroes of f(x), it follows that each one of (x-\sqrt{3}) and (x+\sqrt{3}) is a factor of \mathrm{f}(\mathrm{x})

Consequently, (x-\sqrt{3})(x+\sqrt{3})=\left(x^{2}-3\right) is a factor of f(x).

On dividing f(x) by \left(x^{2}-3\right), we get:

\mathrm{f}(\mathrm{x})=0

\Rightarrow\left(x^{2}+x-20\right)\left(x^{2}-3\right)=0

\Rightarrow\left(x^{2}+5 x-4 x-20\right)\left(x^{2}-3\right)

\Rightarrow[x(x+5)-4(x+5)]\left(x^{2}-3\right)

\Rightarrow(x-4)(x+5)(x-\sqrt{3})(x+\sqrt{3})=0

\Rightarrow \mathrm{x}=4 or \mathrm{x}=-5 or \mathrm{x}=\sqrt{3} or \mathrm{x}=-\sqrt{3}

Hence, all the zeroes are \sqrt{3},-\sqrt{3}, 4 and -5.

i

More Material