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Find the zeroes of the following quadratic polynomials and verify the relationship
Between the zeroes and the coefficients.
$
  \left( {\text{i}} \right){x^2} - 2x - 8{\text{ }}\left( {{\text{ii}}} \right)4{s^2} - 4s + 1\left( {{\text{iii}}} \right)6{x^2} - 3 - 7x \\
  \left( {{\text{iv}}} \right)4{u^2} + 8u\left( {\text{v}} \right){t^2} - 15\left( {{\text{vi}}} \right)3{x^2} - x - 4 \\
 $

Answer
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Hint:-To find zeros of quadratic polynomial first you have to make it in factor form and then you can find it’s zeroes and to verify the relationship between the zeroes and the coefficients use sum of zeroes is $\frac{{ - {\text{b}}}}{a}$and product of zeroes $\frac{{\text{c}}}{a}$.

$\left( {\text{i}} \right){x^2} - 2x - 8$
To convert in factor form we will write it as
$
  {x^2} - 4x + 2x - 8 = 0 \\
  x\left( {x - 4} \right) + 2\left( {x - 4} \right) = 0 \\
  \left( {x - 4} \right)\left( {x + 2} \right) = 0 \\
 $
Now this equation is in it’s factor form so,
$x = 4,x = - 2$
Now we have to verify the relationship between the zeroes and the coefficients.
Sum of zeroes is equal to $\frac{{ - b}}{a}$ on comparing with $a{x^2} + bx + c = 0$, we get $\left( {b = - 2,a = 1} \right)$
That means $\frac{{ - b}}{a} = 2$and sum of zeroes $\left( {4 + \left( { - 2} \right) = 2} \right)$
$\because $Both are equal, hence the relationship is verified.
Now, product of zeroes is equal to $\frac{c}{a}$$\left( {\because c = - 8,a = 1} \right)$
Product of zeroes is $\left( {4 \times - 2 = - 8} \right)$same as $\frac{{ - c}}{a} = - 8$, both are equal hence verified.
$\left( {{\text{ii}}} \right)4{s^2} - 4s + 1 = 0$
To convert it in factor form we will write it as
$
  4{s^2} - 2s - 2s + 1 = 0 \\
  2s\left( {2s - 1} \right) - 1\left( {2s - 1} \right) = 0 \\
  \left( {2s - 1} \right)\left( {2s - 1} \right) = 0 \\
 $
Now this is a factor form so we can easily find $s$from here
$s = \frac{1}{2}$ here both zeroes are same that is $\frac{1}{2}$
Now we have to verify the relationship between zeroes and the coefficients.
Here $\frac{{ - b}}{a} = \frac{4}{4} = 1$and sum of zeroes is $\frac{1}{2} + \frac{1}{2} = 1$ both are equal hence verified.
Here $\frac{c}{a} = \frac{1}{4}$and product of zeroes is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$both are equal hence verified.
$\left( {{\text{iii}}} \right)6{x^2} - 3 - 7x = 0$
Now we have to convert it in factor form, so we will write it as
$
  6{x^2} - 9x + 2x - 3 = 0 \\
  3x\left( {2x - 3} \right) + 1\left( {2x - 3} \right) = 0 \\
  \left( {2x - 3} \right)\left( {3x + 1} \right) = 0 \\
 $
So $x = \frac{3}{2},x = \frac{{ - 1}}{3}$
Now we have to verify the relationship between zeroes and coefficients.
Here $\frac{{ - b}}{a} = \frac{7}{6}$and sum of zeroes $\frac{3}{2} + \frac{{ - 1}}{3} = \frac{7}{6}$both are equal hence verified.
Here $\frac{c}{a} = \frac{{ - 1}}{2}$and product of zeroes $\frac{3}{2} \times \frac{{ - 1}}{3} = \frac{{ - 1}}{2}$both are equal hence verified.
$\left( {{\text{iv}}} \right)4{u^2} + 8u = 0$
Now we have to convert it in factor form
$
  4u\left( {u + 2} \right) = 0 \\
  \therefore u = - 2,u = 0 \\
 $
Here $\frac{{ - b}}{a} = - 2$and sum of zeroes is $ - 2 + 0 = - 2$both are the same hence verified.
Here $\frac{c}{a} = 0$and product of zeroes is $ - 2 \times 0 = 0$both are the same hence verified.
$\left( v \right){t^2} - 15 = 0$
We have to convert it in factor form
$
  {t^2} - {\left( {\sqrt {15} } \right)^2} = 0 \\
  \left( {t - \sqrt {15} } \right)\left( {t + \sqrt {15} } \right) = 0 \\
 $
$t = \sqrt {15} ,t = \sqrt { - 15} $
Here $\frac{{ - b}}{a} = 0$ and sum of zeroes $\sqrt {15} - \sqrt {15} = 0$both are the same hence verified.
Here $\frac{c}{a} = - 15$and product of zeroes $\sqrt {15} \times \sqrt { - 15} = - 15$both are the same hence verified.
$\left( {{\text{vi}}} \right)3{x^2} - x - 4$
We will convert it in factor form
$
  3{x^2} - x - 4 = 0 \\
  3{x^2} - 4x + 3x - 4 = 0 \\
  3x\left( {x + 1} \right) - 4\left( {x + 1} \right) = 0 \\
  \left( {x + 1} \right)\left( {3x - 4} \right) = 0 \\
  x = - 1,x = \frac{4}{3} \\
$
Here $\frac{{ - b}}{a} = \frac{1}{3}$and sum of zeroes $\frac{4}{3} - 1 = \frac{1}{3}$both are same hence verified.
Here $\frac{c}{a} = \frac{{ - 4}}{3}$and product of zeroes $\frac{4}{3} \times - 1 = \frac{{ - 4}}{3}$ both are same hence verified.

Note:-Whenever you get this type of question the key concept of solving is first you have to make a factor form using you basic mathematics and then using properties of quadratic equation you have to check sum of roots or zeros or product of roots with coefficients of polynomial.