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What is the HCF of \(x^4 - 13x^2y^2 - 300y^4\), \(x^3 - 4x^2y - 4xy^2 - 5y^3\) and \(x^3 - 125y^3\)?

This question was previously asked in
CDS 2 2024 Maths Question Paper (01-Sep-2024)
The correct answer is
\(x - 5y\)

Finding the HCF of Given Polynomials

The question asks for the Highest Common Factor (HCF) of three given polynomials:

  1. \(P_1 = x^4 - 13x^2y^2 - 300y^4\)
  2. \(P_2 = x^3 - 4x^2y - 4xy^2 - 5y^3\)
  3. \(P_3 = x^3 - 125y^3\)

To find the HCF, we need to factorize each polynomial completely and then identify the common factors.

Factorizing Polynomial 1: \(x^4 - 13x^2y^2 - 300y^4\)

This polynomial can be treated as a quadratic equation in terms of \(x^2\). Let \(u = x^2\). The expression becomes \(u^2 - 13uy^2 - 300y^4\). We need to find two terms whose product is \(-300y^4\) and whose sum is \(-13y^2\).

By trying factors of 300, we find that \(-25y^2\) and \(12y^2\) multiply to \(-300y^4\) and add up to \(-13y^2\).

So, we can rewrite the expression as:

\(u^2 - 25y^2u + 12y^2u - 300y^4\)

Factoring by grouping:

\(u(u - 25y^2) + 12y^2(u - 25y^2)\)

\((u + 12y^2)(u - 25y^2)\)

Substitute back \(u = x^2\):

\((x^2 + 12y^2)(x^2 - 25y^2)\)

The term \(x^2 - 25y^2\) is a difference of squares (\(a^2 - b^2 = (a-b)(a+b)\)), where \(a=x\) and \(b=5y\). So, \(x^2 - 25y^2 = (x - 5y)(x + 5y)\).

Therefore, the complete factorization of the first polynomial is:

\(P_1 = (x - 5y)(x + 5y)(x^2 + 12y^2)\)

Factorizing Polynomial 2: \(x^3 - 4x^2y - 4xy^2 - 5y^3\)

Let's test if \((x - 5y)\) is a factor by substituting \(x = 5y\) into the polynomial:

\((5y)^3 - 4(5y)^2y - 4(5y)y^2 - 5y^3\)

\(= 125y^3 - 4(25y^2)y - 20y^3 - 5y^3\)

\(= 125y^3 - 100y^3 - 20y^3 - 5y^3\)

\(= 125y^3 - 125y^3 = 0\)

Since the result is 0, \((x - 5y)\) is indeed a factor. Now, we perform polynomial division to find the other factor.

Dividing \(x^3 - 4x^2y - 4xy^2 - 5y^3\) by \((x - 5y)\):

        x^2 + xy + y^2
      ____________________
x - 5y | x^3 - 4x^2y - 4xy^2 - 5y^3
     -(x^3 - 5x^2y)
     ____________________
           x^2y - 4xy^2
          -(x^2y - 5xy^2)
          ____________________
                 xy^2 - 5y^3
                -(xy^2 - 5y^3)
                _____________
                       0

Thus, the factorization is:

\(P_2 = (x - 5y)(x^2 + xy + y^2)\)

Factorizing Polynomial 3: \(x^3 - 125y^3\)

This is a difference of cubes, \(a^3 - b^3\), where \(a = x\) and \(b = 5y\). The formula for the difference of cubes is \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\).

Applying the formula:

\(P_3 = (x - 5y)(x^2 + x(5y) + (5y)^2)\)

\(P_3 = (x - 5y)(x^2 + 5xy + 25y^2)\)

Identifying the Highest Common Factor (HCF)

Now, let's list the factors of each polynomial:

  • \(P_1 = (x - 5y)(x + 5y)(x^2 + 12y^2)\)
  • \(P_2 = (x - 5y)(x^2 + xy + y^2)\)
  • \(P_3 = (x - 5y)(x^2 + 5xy + 25y^2)\)

The common factor present in all three polynomials is \((x - 5y)\).

Therefore, the HCF of the three given polynomials is \(x - 5y\).

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Important Questions from LCM and HCF

  1. The HCF and LCM of two numbers are 12 and 72, respectively. If the ratio of the two numbers is 2 ∶ 3, then the larger of the two numbers is:

  2. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

  3. Joseph visits the club on every 5 th day, Harsh visits on every 24 th day, while Sumit visits on every 9 th day. If all three of them met at the club on a Sunday, then on which day will all three of them meet again?

  4. What is the least number which when divided by 12,20 and 24 leaves in each case a remainder of 8?

  5. Which of the following is a pair of co-primes?

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