The quotient of 8x 3– y 3when divided by 2xy + 4x 2+ y 2is
2x – y
Finding the Quotient of Polynomial Division
The question asks us to find the quotient when the polynomial \(8x^3 - y^3\) is divided by the polynomial \(2xy + 4x^2 + y^2\). This is a problem involving polynomial division, specifically recognizing and applying algebraic identities.
Understanding the Problem
We are given a dividend, \(8x^3 - y^3\), and a divisor, \(2xy + 4x^2 + y^2\). We need to find the expression that results from dividing the dividend by the divisor. Let's rearrange the divisor into a standard form, typically ordered by descending powers of one variable (e.g., \(x\)): \(4x^2 + 2xy + y^2\).
- Dividend: \(8x^3 - y^3\)
- Divisor: \(4x^2 + 2xy + y^2\)
Recognizing Algebraic Identities
The dividend \(8x^3 - y^3\) looks similar to the form of a difference of cubes, which has the algebraic identity:
\[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]Let's see if we can express \(8x^3 - y^3\) in the form \(a^3 - b^3\).
- We can write \(8x^3\) as \((2x)^3\). So, we can let \(a = 2x\).
- We can write \(y^3\) as \((y)^3\). So, we can let \(b = y\).
Now, let's substitute \(a=2x\) and \(b=y\) into the factored form of the difference of cubes identity:
\[(a - b)(a^2 + ab + b^2) = (2x - y)((2x)^2 + (2x)(y) + y^2)\] \[= (2x - y)(4x^2 + 2xy + y^2)\]So, we have factored the dividend \(8x^3 - y^3\) as \((2x - y)(4x^2 + 2xy + y^2)\).
Performing the Division
Now the division becomes:
\[\frac{8x^3 - y^3}{4x^2 + 2xy + y^2} = \frac{(2x - y)(4x^2 + 2xy + y^2)}{4x^2 + 2xy + y^2}\]Assuming the divisor \(4x^2 + 2xy + y^2\) is not equal to zero, we can cancel out the common factor \(4x^2 + 2xy + y^2\) from the numerator and the denominator.
\[\frac{(2x - y)\cancel{(4x^2 + 2xy + y^2)}}{\cancel{4x^2 + 2xy + y^2}} = 2x - y\]Final Quotient
The result of the division is \(2x - y\). This is the quotient.
Verifying with Options
Let's compare our result with the given options:
- \(2x + y\)
- \(x + 2y\)
- \(2x - y\)
- \(4x - y\)
Our calculated quotient \(2x - y\) matches option 3.
| Expression | Type |
|---|---|
| \(8x^3 - y^3\) | Difference of Cubes (Dividend) |
| \(4x^2 + 2xy + y^2\) | Trinomial (Divisor) |
| \(2x - y\) | Binomial (Quotient) |
Revision Table: Polynomial Division & Identities
Understanding polynomial division and recognizing common algebraic identities are key skills. The difference of cubes identity used here is fundamental.
| Identity | Formula | Example Application |
|---|---|---|
| Difference of Cubes | \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) | \(x^3 - 8 = (x-2)(x^2+2x+4)\) |
| Sum of Cubes | \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\) | \(x^3 + 27 = (x+3)(x^2-3x+9)\) |
| Difference of Squares | \(a^2 - b^2 = (a - b)(a + b)\) | \(x^2 - 9 = (x-3)(x+3)\) |
Additional Information: Algebraic Factorization
Algebraic factorization is the process of breaking down an expression into a product of simpler expressions, called factors. Recognizing patterns like the difference or sum of cubes is crucial for efficient factorization, which in turn simplifies expressions and helps in solving equations or performing divisions as shown in this problem.
When dividing polynomials, if the dividend can be factored and one of the factors is exactly the divisor, the quotient is the other factor. If not, polynomial long division or synthetic division might be needed, but for standard problems like this, recognizing identities is the intended method.
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