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Polynomials — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsPolynomials5 Marks
Question
Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$2\text{x}^2 +\frac{7}{2}\text{x}+\frac{3}{4}$.
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Answer

Let $\text{f}\text{(x)}=2\text{x}^2+\frac{7}{2}\text{x}+\frac{3}{4}=8\text{x}^2+14\text{x}+3$
$= 8x^2 + 12x + 2x + 3$ [by splitting the middle term]
$= 4x(2x + 3) + 1(2x + 3)$
$= (2x + 3)(4x + 1)$
So, the value of $8x^2 + 14x + 3$ is zero when 2x + 3 = 0 or 4x + 1 = 0,
i.e., when $\text{x}=-\frac{3}{2}$ or $\text{x}=-\frac{1}{4}$
So, the zeroes of $8x^2 + 14x + 3$ are $-\frac{3}{2}$ and $-\frac{1}{4}$
$\therefore\ \text{Sum of zeroes}=-\frac{3}{2}-\frac{1}{4}=-\frac{7}{4}=\frac{-7}{2\times2}$
$=-\frac{(\text{Coefficient of x})}{(\text{Coefficient of x}^2)}$
and $\text{product of zeroes}=\Big(-\frac{3}{2}\Big)\Big(-\frac{1}{4}\Big)=\frac{3}{8}=\frac{3}{2\times4}$
$=\frac{\text{Constant term}}{\text{Coefficient of x}^2}$
Hence, verified the relations between the zeroes and the coefficients of the polynomial.

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