Relationship between H.C.F & L.C.M

Relationship between H.C.F and L.C.M: Suppose there are two algebraic expressions, $p(x)$ and $q(x)$. The product of H.C.F and L.C.M of $p(x)$ and $q(x)$ equals the product of $p(x)$ and $q(x)$. Mathematically,

$\;\;\;\;\;\;\;\boxed{\text{H.C.F} \times \text{L.C.M} = p(x) \times q(x)}$

Example: Find the H.C.F and L.C.M of $p(x)$ and $q(x)$ given below, and verify the relation of H.C.F and L.C.M. $p(x) = x^2-5x+ 6$; $q(x) = x^2 - 9.$

Solution:

To find the H.C.F and L.C.M of $p(x) = x^2-5x+6$ and $q(x) = x^2-9$, we will first factorize each polynomial.

$p(x) = x^2-5x+6 = (x-2)(x-3)$ $q(x) = x^2-9 = (x+3)(x-3)$

Therefore, we have:

$p(x) = (x-2)(x-3)$ $q(x) = (x+3)(x-3)$

To find the H.C.F of $p(x)$ and $q(x)$, we need to find the common factors of the two polynomials. We can see that the only common factor of the two polynomials is $(x-3)$.

Therefore, the H.C.F of $p(x)$ and $q(x)$ is $(x-3)$.

To find the L.C.M of $p(x)$ and $q(x)$, let's figure out the common and non-common factors.

Common factors $=(x-3)$

Non common factors $= (x+3)(x-2)$

And since, L.C.M = common factors $\times$ non common factors,

therefore, the L.C.M of $p(x)$ and $q(x)$ is $(x-3)(x+3)(x-2)$.

Therefore, the H.C.F of $p(x)$ and $q(x)$ is $(x-3)$ and the L.C.M of $p(x)$ and $q(x)$ is $(x-2)(x-3)(x+3)$.

Now, we can verify the relation between H.C.F and L.C.M, which states that the product of H.C.F and L.C.M of two expression/polynomials is equal to their product. In this case, the two polynomials are $p(x)$ and $q(x)$. Therefore, we have:

$\text{H.C.F}\times\text{L.C.M} = p(x) \times q(x)$

$(x-3)(x-2)(x-3)(x+3) = (x-2)(x-3)(x+3)(x-3)$

$(x-3)^2(x-2)(x+3) = (x-2)(x-3)(x+3)(x-3)$

$(x-3)^2(x-2)(x+3) = (x-2)(x-3)^2(x+3)$

The above equation shows that the relation between the H.C.F and L.C.M of two numbers is true for polynomials/expressions.

Example: Find the L.C.M of $x^3-6x^2+11x-6$ and $x^3-4x+3$, given their H.C.F is $x-1$.

Solution: Given the H.C.F of $x^3-6x^2+11x-6$ and $x^3-4x+3$ is $x-1$, and we know that, $\text{L.C.M} \times \text{H.C.F} =$ $(x^3-6x^2+11x-6)\times (x^3-4x+3)$

But, $\text{H.C.F} = x-1$

$\implies \text{L.C.M} = \frac{(x^3-6x^2+11x-6)\times (x^3-4x+3)}{x-1}$ $= x^3-6x^2+11x-6 \times \frac{x^3-4x+3}{x-1}$

Since,

Hence, the L.C.M is $(x^3-6x^2+11x-6)(x^2+x-3)$.

Solving real-life problems related to H.C.F and L.C.M:

The concepts of LCM and HCF have numerous applications in real-life problems across various fields, including mathematics, science, engineering, finance, and more. For example,

Example: Rida has two pieces of cloth one piece is 45 inches, and the other piece is 90 inches wide. She wants to cut both the strips of equal width. How wide should she cut the strips?

Solution:

To find the width of the strips that Rida should cut, we need to find the HCF (highest common factor) of $45$ and $90$.

We can use the prime factorization method to find the HCF:

$45 = 3 \times 3 \times 5$ $90 = 2 \times 3 \times 3 \times 5$

The common factors are $3^2$ and $5$.

The highest common factor is the product of the common factors:

$\text{HCF} = 3^2 \times 5 = 45$

Therefore, Rida should cut the strips of width $45$ inches to get equal strips from both pieces of cloth.

Example: Sarfraz exercises every $8$ days and Imran every $4$ days. Sarfraz and Imran both exercise today. After how many days do they exercise together again?

Solution:

Since Sarfraz exercises every $8$ days and Imran exercises every $4$ day, we need to find the LCM (lowest common multiple) of $8$ and $4$ to determine the number of days until they both exercise on the same day again.

We can use the prime factorization method to find the LCM:

$8 = 2 \times 2 \times 2$

$4 = 2 \times 2$

Since, LCM = Common Factors $\times$ Non Common Factors

Therefore,

$\text{LCM} = 2 \times 2 \times 2 = 8$

Therefore, Sarfraz and Imran will exercise together again after $8$ days.