Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
\( 4 x^{2}-3 x-1 \)
Given:
$4 x^{2}-3 x-1$
To find:
Here, we have to find the zeros of the given polynomial.
Solution:
Let $f(x)=4x^2 - 3x - 1$
To find the zeros of f(x), we have to put $f(x)=0$.
This implies,
$4x^2 - 3x - 1 = 0$
$4x^2 - 4x + x - 1 = 0$
$4x(x -1 ) + (x - 1) = 0$
$(4x + 1)(x - 1) = 0$
$4x+1=0$ and $x-1=0$
$x = \frac{-1}{4}$ and $x = 1$
Therefore, the zeros of the quadratic equation $f(x) = 4x^2 - 3x - 1$ are $\frac{-1}{4}$ and $1$.
Verification:
We know that,
Sum of zeros $= -\frac{\text { coefficient of x }}{\text { Coefficient of } \mathrm{x}^{2}}$
$= -\frac{(-3)}{4}$
$=\frac{3}{4}$
Sum of the zeros of $f(x)=\frac{-1}{4}+1$
$=\frac{4-1}{4}$
$=\frac{3}{4}$
Product of roots $= \frac{\text { constant term }}{\text { Coefficient of } \mathrm{x}^{2}}$
$= \frac{-1}{4}$
$= -\frac{1}{4}$
Product of the roots of $f(x)=\frac{-1}{4}\times1 =-\frac{1}{4}$
Hence, the relationship between the zeros and their coefficients is verified.
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