3 Marks Question

Question 13 Marks

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$3x^2 + 4x – 4.$

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.

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Question 23 Marks

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$5t^2 + 12t + 7.$

Answer

Let $f(t) = 5t^3 + 12t + 7$
$= 5t^2 + 7t + 5t + 7$ [by splitting the middle term]
$= t(5t + 7) + 1(5t + 7)$
$= (5t + 7)(t + 1)$
Sp, the value of $5t^2 + 12t + 7$ is zero when $5t + 7 = 0$ of $t + 1 = 0,$
i.e., when $\text{t}=\frac{-7}{5}$ of $t = -1,$
So, the zeroes of $5t^2 + 12t + 7$ are $\frac{-7}{5}$ and -1.
$\therefore\ \text{Sum of zeroes}= -\frac{7}{5}-1=\frac{-12}{5}$
$=(-1),\Big(\frac{\text{Coefficient of t}}{\text{Coefficient of t}^2}\Big)$
and $\text{product of zeroes}=-\frac{7}{5}(-1)=\frac{7}{5}$
$=(-1)^2\Big(\frac{\text{Constant term}}{\text{Coefficient of t}^2}\Big)$
Hence, verified the relations between the zeroes and the coefficients of the polynomial.

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