Question
Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$3x^2 + 4x – 4.$
$3x^2 + 4x – 4.$
✓
Answer
Let $f(x) = 3x^2 + 4x - 4= 3x^2 + 6x - 2x - 4$ [ by splitting the middle term]
$= 3x(x + 2) - 2 (x + 2)$
$= (x + 2)(3x - 2)$
So, the value of $3x^2 + 4x - 4$ is zero when $x + 2 = 0$ or $3x - 2 = 0$
i.e.,when x = -2 or $\text{x}=\frac{2}{3}$ So,
the zeroes of $3x2 + 4x - 4$ are $-2$ and $\frac{2}{3}$.
$\therefore\ \text{Sum of zeroes} = -2 +\frac{2}{3}=-\frac{4}{3}$
$=(-1)\Big(\frac{\text{Coefficient of x}}{\text{Coefficient of x}^2}\Big)$
and $\text{product of zeroes}=(-2)\Big(\frac{2}{3}\Big)=\frac{-4}{3}$
$=(-1)^2\Big(\frac{\text{Constant term}}{\text{Coefficient of x}^2}\Big)$
Hence, verified the relations between the zeroes and the coeffiecients of the polynomial.
$= 3x(x + 2) - 2 (x + 2)$
$= (x + 2)(3x - 2)$
So, the value of $3x^2 + 4x - 4$ is zero when $x + 2 = 0$ or $3x - 2 = 0$
i.e.,when x = -2 or $\text{x}=\frac{2}{3}$ So,
the zeroes of $3x2 + 4x - 4$ are $-2$ and $\frac{2}{3}$.
$\therefore\ \text{Sum of zeroes} = -2 +\frac{2}{3}=-\frac{4}{3}$
$=(-1)\Big(\frac{\text{Coefficient of x}}{\text{Coefficient of x}^2}\Big)$
and $\text{product of zeroes}=(-2)\Big(\frac{2}{3}\Big)=\frac{-4}{3}$
$=(-1)^2\Big(\frac{\text{Constant term}}{\text{Coefficient of x}^2}\Big)$
Hence, verified the relations between the zeroes and the coeffiecients of the polynomial.
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