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Factor of Polynomials

Find the zero...

Question

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^{2} - 2 x - 8$
(ii) $4 s^{2} - 4 s + 1$
(iii) $6 x^{2} - 3 - 7 x$
(iv) $4 u^{2} + 8 u$
(v) $t^{2} - 15$
(vi) $3 x^{2} - x - 4$

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Solution

(i) $x^{2} - 2 x - 8$
$x^{2} - 2 x - 8$
$= x^{2} - 4 x + 2 x - 8$
$= x (x - 4) + 2 (x - 4)$
$= (x + 2) (x - 4)$
Factorize the equation, we got $(x + 2) (x - 4)$
So, the value of $x^{2} - 2 x - 8$ is zero when $x + 2 = 0, x - 4 = 0,$ i.e., when $x = - 2$ or $x = 4$ .
Therefore, the zeros of $x^{2} - 2 x - 8$ are -2 and 4.
Now,
$\Rightarrow$ Sum of zeroes $= - 2 + 4 = 2$ = $- \frac{2}{1} = - \frac{C o e f f i c i e n t o f x}{C o e f f i c i e n t o f x^{2}}$

$\Rightarrow$ Product of zeros $= (- 2) \times (4) = - 8$ = $\frac{- 8}{1} = \frac{C o n s t a n t t e r m}{C o e f f i c i e n t o f x^{2}}$

(ii) $4 s^{2} - 4 s + 1$
$4 s^{2} - 4 s + 1$
$= 4 s^{2} - 2 s - 2 s + 1$
$= 2 s (2 s - 1) - 2 (2 s - 1)$
$= (2 s - 1) (2 s - 1)$
Factorize the equation, we got $(2 s - 1) (2 s - 1)$
So, the value of $4 s^{2} - 4 s + 1$ is zero when $2 s - 1 = 0, 2 s - 1 = 0$ , i.e., when $s = \frac{1}{2}$ or $s = \frac{1}{2}$ .
Therefore, the zeros of $4 s^{2} - 4 s + 1$ are $\frac{1}{2}$ and $\frac{1}{2}$ .
Now,
$\Rightarrow$ Sum of zeroes = $\frac{1}{2} + \frac{1}{2} = 1 = - \frac{- 4}{4} = - \frac{C o e f f i c i e n t o f s}{C o e f f i c i e n t o f s^{2}}$

$\Rightarrow$ Product of zeros = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{1}{4} = \frac{C o n s t a n t t e r m}{C o e f f i c i e n t o f s^{2}}$

(iii) $6 x^{2} - 3 - 7 x$
$= 6 x^{2} - 7 x - 3$
$= 6 x^{2} - 9 x + 2 x - 3$
$= 3 x (2 x - 3) + 1 (2 x - 3)$
$= (3 x + 1) (2 x - 3)$
Factorize the equation, we got $(3 x + 1) (2 x - 3)$
So, the value of $6 x^{2} - 3 - 7 x$ is zero when $3 x + 1 = 0, 2 x - 3 = 0$ , i.e., when $x = - \frac{1}{3}$ or $x = \frac{3}{2}$ .
Therefore, the zeros of $6 x^{2} - 3 - 7 x$ are $- \frac{1}{3}$ and $\frac{3}{2}$ .
Now,
$\Rightarrow$ Sum of zeroes = $- \frac{1}{3} + \frac{3}{2} = \frac{7}{6} = - \frac{- 7}{6} = - \frac{C o e f f i c i e n t o f x}{C o e f f i c i e n t o f x^{2}}$

$\Rightarrow$ Product of zeros = $- \frac{1}{3} \times \frac{3}{2} = - \frac{1}{2} = \frac{- 3}{6} = \frac{- 1}{2} = \frac{C o n s t a n t t e r m}{C o e f f i c i e n t o f x^{2}}$

(iv) $4 u^{2} + 8 u$
$= 4 u (u + 2)$
Factorize the equation, we got $4 u (u + 2)$
So, the value of $4 u^{2} + 8 u$ is zero when $4 u = 0, u + 2 = 0$ , i.e., when $u = 0$ or $u = - 2$ .
Therefore, the zeros of $4 u^{2} + 8 u$ are $0$ and $- 2.$
Now,
$\Rightarrow$ Sum of zeroes = $0 - 2 = - 2 = - \frac{8}{4} = - 2 = - \frac{C o e f f i c i e n t o f u}{C o e f f i c i e n t o f u^{2}}$

$\Rightarrow$ Product of zeros = $- 0 x - 2 = 0 = \frac{0}{4} = 0 = \frac{C o n s t a n t t e r m}{C o e f f i c i e n t o f u^{2}}$

(v) $t^{2} - 15$
$= t^{2} - (\sqrt{15})^{2}$
$= (t + \sqrt{15}) (t - \sqrt{15})$
So, the value of $t^{2} - 15$ is zero when $t + \sqrt{15} = 0, t - \sqrt{15} = 0$ , i.e., when $t = \sqrt{15}$ or $t = - \sqrt{15}$ .
Therefore, the zeros of $t^{2} - 15$ are $\pm \sqrt{15}$ .
Now,
$\Rightarrow$ Sum of zeroes = $\sqrt{15} - \sqrt{15} = 0 = - \frac{0}{1} = 0 = - \frac{C o e f f i c i e n t o f t}{C o e f f i c i e n t o f t^{2}}$

$\Rightarrow$ Product of zeros = $\sqrt{15} \times \sqrt{- 15} = - 15 = \frac{- 15}{1} = \frac{C o n s t a n t t e r m}{C o e f f i c i e n t o f t^{2}}$

(vi) $3 x^{2} - x - 4$
$= 3 x^{2} - x - 4$
$= 3 x^{2} + 3 x - 4 x - 4$
$= 3 x (x + 1) - 4 (x + 1)$
$= (x + 1) (3 x - 4)$
Factorize the equation, we get $(x + 1) (3 x - 4)$
So, the value of $3 x^{2} - x - 4$ is zero when x + 1 = 0, 3x - 4 = 0, i.e., when x = -1 or x = $\frac{4}{3}$ .
Therefore, the zeros of $3 x^{2} - x - 4$ are -1 and $\frac{4}{3}$ .
Now,
$\Rightarrow$ Sum of zeroes = $- 1 + \frac{4}{3} = \frac{1}{3} = - \frac{- 1}{3} = - \frac{C o e f f i c i e n t o f x}{C o e f f i c i e n t o f x^{2}}$

$\Rightarrow$ Product of zeros = $- 1 \times \frac{4}{3} = - \frac{4}{3} = \frac{- 4}{3} = \frac{C o n s t a n t t e r m}{C o e f f i c i e n t o f x^{2}}$

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Question 1 (iv)
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

$4 u^{2} + 8 u$

4 u^{2} + 8 u

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