Find the lowest common multiple (LCM) of two or more algebraic expressions; examples are presented along with their detailed solutions and questions with solutions and detailed explanations are also included.
The lowest common multiple of two or more expressions is the smallest (or simplest) expression that is divisible by each of these expressions. It is found by first factoring completely each of the given expressions then use these factors to write the LCM. Detailed examples are shown below.
\( \) \( \)\( \)\( \) Find the lowest common multiple of the two expressions: \(x^2 - 1\) and \(x - 1\).
\(x - 1\) is a factor to both expression and will therefore be used once. \(x + 1\) is a factor in the first expression and will therefore be used. Hence \[ \text{LCM of} \; ( x^2 - 1 \quad , \quad x - 1 ) = \color{red} {(x - 1)(x+1)} \]
Find the lowest common multiple of the three expressions: \(2 x^2\) , \( x^2 + x \) and \(x^3 + 2 x \).
\(2\) is a factor in the first term only and will therefore be used. \(x\) is a factor in all three expressions and the one with the highest power which is \(x^2\) in the first term is used. \(x + 1\) is a factor in the second expression only and is therefore used. \(x^2 + 1\) is a factor in the third expression only and is therefore used. Hence \[ \text{LCM of} \; ( 2 x^2 \quad , \quad x^2 + x \quad , \quad x^3 + 2 x ) = \color{red} {2x^2 (x + 1) (x^2 + 2)} \]
\(x - 1\) is a factor in the first and second expressions is therefore the one with the highest power \((x - 1)^2\) in the second expression is used. \(x + 4\) is a factor in the first and third expressions is used once only. \(x + 5\) is a factor in the third expression only and is therefore used . Hence \[ \text{LCM of} \; ( x^2 + 3 x - 4 \quad , \quad (x - 1)^2 \quad , \quad x^2 + 9 x + 20 ) = \color{red} {(x - 1)^2 (x + 4)(x + 5)} \]
Find Lowest Common Multiple (LCM) of the algebraic expressions given below.
\(2\) is a factor in the first expressions is therefore used. \(x + 1\) is a factor in the first and second expressions is used once only. \(3\) is a factor in the second expression only and is therefore used . Hence \[ \text{LCM of} \; ( 2 (x + 1) \quad , \quad 3 (x + 1) ) = \color{red} { 2 \cdot 3 (x + 1) = 6 (x+1) } \]
\(2\) is a factor in the first expressions is therefore used. \(x - 1\) is a factor in the first and second expressions and the factor with the highest power which is \((x - 1)^2\) in the first expression is used. \(5\) is a factor in the second expression only and is therefore used. Hence \[ \text{LCM of} \; ( 2 (x - 1)^2 \quad , \quad 5 (x - 1) ) = \color{red} { 2 \cdot 5 (x - 1)^2 = 10 (x - 1)^2 } \]
\(x + 3\) is a factor in the first expressions is therefore used. \(x + 2\) is a factor in the first and second expressions and therefore used once. \(2\) is a factor in the second expression only and is therefore used. \(x - 1\) is a factor in the second expression and is therefore used. Hence
\(x\) is a factor in the first expressions is therefore used. \(3 x + 1\) is a factor in the first expression and is therefore used. \(x - 1\) is a factor in the first and second expression and is therefore used once only. Hence
We now make the LCM by multiplying all factors included in the factoring of the given expressions. Common factors are used once only and the one with the highest power is used.
\(x\) is a factor in the first expressions is therefore used. \(3 x + 1\) is a factor in the first expression and is therefore used. \(x - 1\) is a factor in all three expressions and the one with the highest power which is \((x - 1)^2\) in the third expression is used. \(2\) is a factor in the second expression and is therefore used. \(x + 1\) is a factor in the second expression and is therefore used. Hence