3. Obtain all other zeroes of $3x^4 + 6x^3 – 2x^2 – 10x – 5$, if two of its zeroes are
${√{5/3}}$ and ${-√{5/3}}$
Solution:
Since ${√{5/3}}$ and ${-√{5/3}}$ are its zeroes,
therefore $(x- {√{5/3})(x+{√{5/3})$ are factor of the polynomial.
i.e $x^2-{5/3}$ or $3x^2-5$ is factor of the polymial.
Divide $3x^4 + 6x^3 – 2x^2 – 10x – 5$ by $3x^2-5$
$3x^4 + 6x^3 – 2x^2 – 10x – 5$
${3x^2-5}$
$x^2$
$+ 2x$
$+1$
${3x^4}$ ${-5x^2}$($-$)
($+$)
$6x^3 + 3x^2 -10x - 5 $
${6x^3 }$ ${-10x}$($-$)
($+$)
${3x^2}$ ${-5}$${3x^2}$ ${-5}$($-$) ($+$)
0
Since the remainder is zero, $x^2+2x+1$ is a factor of $3x^4 + 6x^3 – 2x^2 – 10x – 5$
Factorised $x^2+2x+1$ by splitting the middle term
Therefore $x^2+2x+1$ = $x^2+x+x+1$ = $x(x+1)+1(x+1)$ =$(x+1)(x+1)$ So zeores of the polynomial are $-1$ and $-1$