Question

The number of zeroes of polynomial f(x) = (x - 1)(x-2)(x-3) are (a) 1 (c) 3 (6) 2 (d) 4
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Answer 1

X-1 =0
X=1

X-2=0
X=2

X-3=0
X=3

Answer 2

Question:

The number of zeroes of polynomial f(x) = (x - 1)(x-2)(x-3) are

(a) 1

(b) 3

(c) 2

(d) 4

Given:

f(x) = (x - 1) (x - 2) (x - 3)

Solution:

First Multiplying (x - 1) with (x - 2)

$\tt \large \:⇒ (x - 1) \times (x - 2)$

$\tt \small \:⇒ {x}^{2} + (x \times - 2) +( - 1 \times x) + ( - 1\times - 2)$

$\tt \large \:⇒ {x}^{2} - 2x - 1x + 2$

$\tt \large \:⇒ {x}^{2} - 3x+ 2$

Now, multiplying

$\tt ({x}^{2} - 3x+ 2) \: with \: (x - 3)$

$\tt \large \:⇒ ({x}^{2} - 3x+ 2) \times (x - 3)$

$\tt \footnotesize \:⇒ {x}^{3} + ({x}^{2}\times - 3) +( - 3x \times x) + ( - 3x\times - 3) + (2 \times x) + ( - 3 \times 2)$

$\tt \large \:⇒ {x}^{3} - 3{x}^{2} - 3 {x}^{2} + 9 {x}^{2} + 2 x - 6$

$\tt \large \:⇒ {x}^{3} + 3{x}^{2} + 2 x - 6$

Here as we can see the highest power of x is 3. So, we can conclude that the number of zeroes of polynomial f(x) = (x - 1)(x-2)(x-3) will also 3.

Henceforth,

Option (b) is the right option.

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