Find zeroes of v²+4√3v-15Also verify relationship …

The given quadratic polynomial is:

  v² + 4$\sqrt 3$v - 15

By factorizing it we have
   v² + 4$\sqrt 3$v - 15  = v² + 5$\sqrt 3$v - $\sqrt 3$v - 15
= v(v + 5$\sqrt 3$) - $\sqrt 3$(v + 5$\sqrt 3$)
 = (v - $\sqrt 3$)(v + 5$\sqrt 3$)

For zeroes, put the factors equal to zero i.e., 

(v - $\sqrt 3$)(v + 5$\sqrt 3$) = 0
$\Rightarrow v = \sqrt { 3 } , - 5 \sqrt { 3 }$ are zeroes of the polynomial.

Verification: In the given polynomial a = 1, b =  4$\sqrt 3$ and c =  - 15
Now Sum of the zeroes = $\sqrt { 3 } + ( - 5 \sqrt { 3 } ) = - 4 \sqrt { 3 }$
Also sum of zeroes = $\frac { - b } { a }$, $\frac { - b } { a } = \frac { - 4 \sqrt { 3 } } { a } = - 4 \sqrt { 3 }$
And product of zeroes = $\sqrt { 3 } \times - 5 \sqrt { 3 }$= -15
Also, product of zeroes = $\frac { c } { a } = \frac { - 15 } { 1 } = - 15$