Zeros of px-x2-2 x-3 area 1-3b 3-1c -3-1d 1-3

(b) 3, 1 Here, p(x)=x2 2x 3 Let #160;x2 2x 3=0= #62;x2 (3 1)x 3=0= #62;x2 3x+x 3=0= #62;xx 3+1x 3=0= #62;x 3x+1=0= #62;x=3, 1 ...

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Standard X

Mathematics

Factor of a Polynomial

Zeros of px=x...

Question

Zeros of p(x) = x² − 2x − 3 are

(a) 1, −3
(b) 3, −1
(c) −3, −1
(d) 1, 3

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Solution

(b) 3,-1
Here, ${p (x) = x}^{2} - 2 x - 3$

$Let x^{2} - 2 x - 3 = 0 = > x^{2} - (3 - 1) x - 3 = 0 = > x^{2} - 3 x + x - 3 = 0 = > x (x - 3) + 1 (x - 3) = 0 = > (x - 3) (x + 1) = 0 = > x = 3, - 1$

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Similar questions

Q. The zeros of the polynomial x² − 2x − 3 are

(a) −3, 1
(b) −3, −1
(c) 3, −1
(d) 3, 1

The zeros of the polynomial $p (x) = 3 x^{2} - 1$ are
(a) $\frac{1}{3}$ and 3
(b) $\frac{1}{\sqrt{3}}$ and $\sqrt{3}$
(c) $\frac{- 1}{\sqrt{3}}$ and $\sqrt{3}$
(d) $\frac{1}{\sqrt{3}}$ and $\frac{- 1}{\sqrt{3}}$

Q. What is the quotient obtained when

x^{3} + x^{2} - 5 x + 3

is divided by

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Q. Find the zeros of each of the following quadratic polynomial and verify the relationship between the zeros and their coefficients :
(i)

f (x) = x^{2} - 2 x - 8

(ii)

g (x) = 4 s^{2} - 4 s + 1

(iii)

f (x) = x^{2} - (\sqrt{3} + 1) x + \sqrt{3}

(iv)

x^{2} - 3 - 7 x

(v)

p (x) = x^{2} + 2 \sqrt{2} x - 6

(vi)

q (x) = \sqrt{3} x^{2} + 10 x + 7 \sqrt{3}

(vii)

g (x) = a (x^{2} + 1) - x (a^{2} + 1)

Q. The line

\frac{x - 2}{4} = \frac{y + 1}{- 1} = \frac{z - 1}{3}

is perpendicular to the plane

(a) 2 x - y + z = 3

(b) 4 x - y + 3 z = 1

(c) 4 x + y + 3 z = 1

(d) x + 7 y + z = 3

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Zeros of p(x) = x2 − 2x − 3 are (a) 1, −3 (b) 3, −1 (c) −3, −1 (d) 1, 3

(b) 3,-1 Here, p(x)=x2-2x-3 Let x2-2x-3=0=>x2-(3-1)x-3=0=>x2-3x+x-3=0=>xx-3+1x-3=0=>x-3x+1=0=>x=3,-1

Zeros of p(x) = x² − 2x − 3 are

(a) 1, −3
(b) 3, −1
(c) −3, −1
(d) 1, 3

(b) 3,-1
Here, ${p (x) = x}^{2} - 2 x - 3$

$Let x^{2} - 2 x - 3 = 0 = > x^{2} - (3 - 1) x - 3 = 0 = > x^{2} - 3 x + x - 3 = 0 = > x (x - 3) + 1 (x - 3) = 0 = > (x - 3) (x + 1) = 0 = > x = 3, - 1$