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

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Prepp Team · Accepted Answer

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: Factorizing each polynomial completely into its irreducible factors over a specified field (usually real numbers or rational numbers in this context). Identifying the factors that are common to all the polynomials. 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 $(x^2 + y^2)$.

Prepp Team · Answer

(x2 + y2)

Prepp Team · Answer

(x2 - y2)

Prepp Team · Answer

(x3- y3+ x2y – xy2)

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

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

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

Identifying Common Factors and Calculating HCF

Comparing with Options

Revision Table: Key Algebraic Factorizations

Additional Information: HCF of Polynomials

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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

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