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Byju's Answer

Standard X

Mathematics

LCM of Polynomials

Find the LCM ...

Question

Find the LCM of the pair of the given polynomials.
$(x - 1) (x - 2) (x^{2} - 3 x + 3), (x - 1), (x^{3} - 4 x^{2} + 6 x - 3)$

(x - 2) (x - 1)

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(x - 1)

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(x - 3) (x - 1)

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(x - 1) (x - 2) (x^{2} - 3 x + 3)

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Solution

The correct option is D $(x - 1) (x - 2) (x^{2} - 3 x + 3)$
We know that the least common multiple (LCM) is the smallest number or expression that is a common multiple of two or more numbers or algebraic terms.

We first factorize the given polynomials as shown below:

$(x - 1) (x - 2) (x^{2} - 3 x + 3) = (x - 1) (x - 2) (x^{2} - 3 x + 3)$
$(x - 1) = (x - 1)$
$(x^{3} - 4 x^{2} + 6 x - 3) = (x - 1) (x^{2} - 3 x + 3)$

Now multiply all the factors, using each common factor only once, therefore, the LCM is:
$= (x - 1) (x - 2) (x^{2} - 3 x + 3)$

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Similar questions

Differentiate each the following from first principles :

$(i) \frac{2}{x}$

$(i i) \frac{1}{\sqrt{x}}$

$(i i i) \frac{1}{x^{3}}$

$(i v) \frac{x^{2} + 1}{x}$

$(v) \frac{x^{2} - 1}{x}$

$(v i) \frac{x + 1}{x + 2}$

$(v i i) \frac{x + 2}{3 x + 5}$

$(v i i i) k x^{n}$

$(i x) \frac{1}{\sqrt{3 - x}}$

$(x) x^{2} + x + 3$

$(x i) (x + 2)^{3}$

$(x i i) x^{3} + 4 x^{2} + 3 x + 2$

$(x i i i) (x^{2} + 1) (x - 5)$

$(x i v) \sqrt{2 x^{2} + 1}$

$(x v) \frac{2 x + 3}{x - 2}$

Q. (i)

\frac{2}{x}

(ii)

\frac{1}{\sqrt{x}}

(iii)

\frac{1}{x^{3}}

(iv)

\frac{x^{2} + 1}{x}

(v)

\frac{x^{2} - 1}{x}

(vi)

\frac{x + 1}{x + 2}

(vii)

\frac{x + 2}{3 x + 5}

(viii) k xⁿ

(ix)

\frac{1}{\sqrt{3 - x}}

(x) x² + x + 3

(xi) (x + 2)³

(xii) x³ + 4x² + 3x + 2

(xiii) (x² + 1) (x − 5)

(xiv)

\sqrt{2 x^{2} + 1}

(xv)

\frac{2 x + 3}{x - 2}

Q. Which of the following are quadratic equations?

(i) x² + 6x − 4 = 0
(ii)

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

(iii)

x^{2} + \frac{1}{x^{2}} = 5

(iv)

x - \frac{3}{x} = x^{2}

(v)

2 x^{2} - \sqrt{3 x} + 9 = 0

(vi)

x^{2} - 2 x - \sqrt{x} - 5 = 0

(vii) 3x² − 5x + 9 = x² − 7x + 3
(viii)

x + \frac{1}{x} = 1

(ix) x² − 3x = 0
(x)

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

(xi) (2x + 1) (3x + 2) = 6(x − 1) (x − 2)
(xii)

x + \frac{1}{x} = x^{2}, x \neq 0

(xiii) 16x² − 3 = (2x + 5) (5x − 3)
(xiv) (x + 2)³ = x³ − 4
(xv) x(x + 1) + 8 = (x + 2) (x − 2)

Quotient and remainder of a polynomial when divided by $x + 2$ are $2 x - 1$ and $3$ respectively. Find the polynomial.

Check whether the following are quadratic equations :
(i) ${(x + 1)}^{2} = 2 (x - 3)$
(ii) $x^{2} - 2 x = (- 2) (3 - x)$
(iii) $(x - 2) (x + 1) = (x - 1) (x + 3)$
(iv) $(x - 3) (2 x + 1) = x (x + 5)$
(v) $(2 x - 1) (x - 3) = (x + 5) (x - 1)$
(vi) $x^{2} + 3 x + 1 = {(x - 2)}^{2}$
(vii) ${(x + 2)}^{3} = 2 x (x^{2} - 1)$
(viii) $x^{3} - 4 x^{2} - x + 1 = {(x - 2)}^{3}$

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Find the LCM of the pair of the given polynomials. (x−1)(x−2)(x2−3x+3),(x−1),(x3−4x2+6x−3)

Find the LCM of the pair of the given polynomials.
$(x - 1) (x - 2) (x^{2} - 3 x + 3), (x - 1), (x^{3} - 4 x^{2} + 6 x - 3)$