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Question

What is the HCF of (x8– y8) and (x7– y7+ x5y2– x2y5) ? 

The correct answer is

(x3 - y3  - x2y  + xy2

Finding the HCF of Polynomials: \(x^8 - y^8\) and \(x^7 - y^7 + x^5y^2 - x^2y^5\)

To find the Highest Common Factor (HCF) of two polynomial expressions, we need to factorize each expression completely and then identify the common factors.

Factorizing the First Polynomial: \(x^8 - y^8\)

The first polynomial is \(x^8 - y^8\). This expression is a difference of squares, as \(x^8 = (x^4)^2\) and \(y^8 = (y^4)^2\). We can apply the difference of squares formula, \(a^2 - b^2 = (a - b)(a + b)\).

Applying the formula:

$$x^8 - y^8 = (x^4)^2 - (y^4)^2$$ $$= (x^4 - y^4)(x^4 + y^4)$$

Now, the term \((x^4 - y^4)\) is also a difference of squares, as \(x^4 = (x^2)^2\) and \(y^4 = (y^2)^2\). We apply the formula again:

$$x^4 - y^4 = (x^2)^2 - (y^2)^2$$ $$= (x^2 - y^2)(x^2 + y^2)$$

The term \((x^2 - y^2)\) is yet another difference of squares: \(x^2 - y^2 = (x - y)(x + y)\).

Substituting back, the complete factorization of \(x^8 - y^8\) is:

$$x^8 - y^8 = (x - y)(x + y)(x^2 + y^2)(x^4 + y^4)$$

Factorizing the Second Polynomial: \(x^7 - y^7 + x^5y^2 - x^2y^5\)

The second polynomial is \(x^7 - y^7 + x^5y^2 - x^2y^5\). We can try grouping terms to find common factors.

Let's group the terms as \((x^7 + x^5y^2) + (-y^7 - x^2y^5)\). Notice that the second group has a negative sign. We can rewrite it as \((x^7 + x^5y^2) - (y^7 + x^2y^5)\).

In the first group, \((x^7 + x^5y^2)\), we can factor out \(x^5\):

$$x^7 + x^5y^2 = x^5(x^2 + y^2)$$

In the second group, \((y^7 + x^2y^5)\), we can factor out \(y^5\):

$$y^7 + x^2y^5 = y^5(y^2 + x^2) = y^5(x^2 + y^2)$$

Now substitute these back into the expression:

$$x^7 - y^7 + x^5y^2 - x^2y^5 = x^5(x^2 + y^2) - y^5(x^2 + y^2)$$

We can now see that \((x^2 + y^2)\) is a common factor in both terms. Factor it out:

$$x^5(x^2 + y^2) - y^5(x^2 + y^2) = (x^5 - y^5)(x^2 + y^2)$$

Now we need to factor \((x^5 - y^5)\). This is a difference of fifth powers. The general formula for \(a^n - b^n\) when n is any positive integer is \((a - b)(a^{n-1} + a^{n-2}b + \dots + ab^{n-2} + b^{n-1})\). For \(n=5\):

$$x^5 - y^5 = (x - y)(x^4 + x^3y + x^2y^2 + xy^3 + y^4)$$

Substituting this back, the complete factorization of the second polynomial is:

$$x^7 - y^7 + x^5y^2 - x^2y^5 = (x - y)(x^4 + x^3y + x^2y^2 + xy^3 + y^4)(x^2 + y^2)$$

Identifying Common Factors and Calculating HCF

Let's list the factors for both polynomials:

  • Factors of \(x^8 - y^8\): \((x - y), (x + y), (x^2 + y^2), (x^4 + y^4)\)
  • Factors of \(x^7 - y^7 + x^5y^2 - x^2y^5\): \((x - y), (x^4 + x^3y + x^2y^2 + xy^3 + y^4), (x^2 + y^2)\)

The common factors are \((x - y)\) and \((x^2 + y^2)\).

The HCF is the product of the common factors:

$$\text{HCF} = (x - y)(x^2 + y^2)$$

Let's expand this expression:

$$(x - y)(x^2 + y^2) = x(x^2 + y^2) - y(x^2 + y^2)$$ $$= x \cdot x^2 + x \cdot y^2 - y \cdot x^2 - y \cdot y^2$$ $$= x^3 + xy^2 - x^2y - y^3$$

Rearranging the terms to match the options, we get \(x^3 - y^3 - x^2y + xy^2\).

Comparing with Options

Let's compare our calculated HCF, \(x^3 - y^3 - x^2y + xy^2\), with the given options:

Option Expression Matches HCF?
1 \((x^2 + y^2)\) No
2 \((x^2 - y^2)\) No
3 \((x^3 - y^3 - x^2y + xy^2)\) Yes
4 \((x^3 - y^3 + x^2y - xy^2)\) No

Our calculated HCF matches option 3.

Revision Table: Key Algebraic Factorizations

Formula Description Example
\(a^2 - b^2 = (a - b)(a + b)\) Difference of Squares \(x^4 - y^4 = (x^2 - y^2)(x^2 + y^2)\)
\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) Difference of Cubes \(x^6 - y^6 = (x^2)^3 - (y^2)^3 = (x^2 - y^2)(x^4 + x^2y^2 + y^4)\)
\(a^n - b^n = (a - b)(a^{n-1} + \dots + b^{n-1})\) Difference of nth Powers \(x^5 - y^5 = (x - y)(x^4 + x^3y + x^2y^2 + xy^3 + y^4)\)

Additional Information: HCF of Polynomials

The Highest Common Factor (HCF), also known as the Greatest Common Divisor (GCD), of two or more polynomials is the polynomial of the highest possible degree that divides each of the given polynomials. Finding the HCF of polynomials is analogous to finding the HCF of numbers.

The process typically involves:

  1. Factorizing each polynomial completely into its irreducible factors over a specified field (usually real numbers or rational numbers in this context).
  2. Identifying the factors that are common to all the polynomials.
  3. Multiplying the common factors, raised to the lowest power they appear in any of the factorized polynomials.

In this problem, we found the HCF by factoring both polynomials and identifying the common irreducible factors \((x - y)\) and \((x^2 + y^2)\).

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Important Questions from LCM and HCF

  1. Six bells begin to toll together and toll, respectively, at intervals of 3, 4, 6, 7, 8 and 12 seconds. After how many seconds, will they toll together again?

  2. A and B are two prime numbers such that A > B and their LCM is 209. The value of A 2 - B is:

  3. Find the least number which when divided by 12, 18, 24 and 30 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.

  4. Calculate the HCF of \(\frac{12}{5}\) \(\frac{14}{15}\)  and  \(\frac{16}{17}\) .

  5. Three numbers are in the proportion of 3 : 8 : 15 and their LCM is 8280. What is their HCF?

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