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