What is the sum of the linear factors (in x and y) of the expression 2x 2 + xy - 3y 2 ?
3x + 2y
Finding the Sum of Linear Factors
The question asks for the sum of the linear factors of the expression $\(2x^2 + xy - 3y^2\)$. This is a quadratic expression involving two variables, $\(x\)$ and $\(y\)$. To find the sum of its linear factors, we first need to factorize the expression into two linear terms.
Factorizing the Expression $\(2x^2 + xy - 3y^2\)$
We can factor this quadratic expression by treating it similarly to a quadratic trinomial in one variable. We look for two linear factors of the form $\((ax + by)(cx + dy)\)$ whose product is $\(2x^2 + xy - 3y^2\)$.
Let's try to factor by splitting the middle term $\(xy\)$. We need two terms whose product is $\(2x^2 \times -3y^2 = -6x^2y^2\)$ and whose sum is $\(xy\)$. The coefficients of these terms must multiply to \(-6\) and add up to \(1\). The numbers are \(3\) and \(-2\). So, we can rewrite $\(xy\)$ as $\(3xy - 2xy\)$.
The expression becomes:
\(2x^2 + xy - 3y^2 = 2x^2 + 3xy - 2xy - 3y^2\)
Now, we group terms and factor by grouping:
\((2x^2 + 3xy) + (-2xy - 3y^2)\)
Factor out the common term from the first group and the second group:
\(x(2x + 3y) - y(2x + 3y)\)
Notice that we now have a common binomial factor, which is $\((2x + 3y)\)$. Factor this out:
\((2x + 3y)(x - y)\)
So, the linear factors of the expression $\(2x^2 + xy - 3y^2\)$ are $\((2x + 3y)\)$ and $\((x - y)\)$.
Calculating the Sum of the Linear Factors
The question asks for the sum of these two linear factors. We need to add $\((2x + 3y)\)$ and $\((x - y)\)$.
Sum = $\((2x + 3y) + (x - y)\)$
Remove the parentheses and combine like terms (terms with $\(x\)$ and terms with $\(y\)$):
Sum = $\(2x + 3y + x - y\)
Group the $\(x\)$ terms and the $\(y\)$ terms:
Sum = $\((2x + x) + (3y - y)\)
Perform the addition and subtraction:
Sum = $\(3x + 2y\)
The sum of the linear factors of the expression $\(2x^2 + xy - 3y^2\)$ is $\(3x + 2y\)$.
Comparing with Options
Let's compare our calculated sum with the given options:
- Option 1: $\(2x - 3y\)$
- Option 2: $\(3x - 2y\)$
- Option 3: $\(3x + 2y\)$
- Option 4: $\(2x + 3y\)$
Our calculated sum $\(3x + 2y\)$ matches Option 3.
| Expression | Factored Form | Linear Factors | Sum of Factors |
|---|---|---|---|
| $\(2x^2 + xy - 3y^2\)$ | $\((2x + 3y)(x - y)\)$ | $\((2x + 3y)\)$, $\((x - y)\)$ | $\((2x + 3y) + (x - y) = 3x + 2y\)$ |
Revision Table: Factoring and Summing Linear Factors
| Concept | Description | Example/Process |
|---|---|---|
| Quadratic Expression in Two Variables | An expression of the form $\(ax^2 + bxy + cy^2 + dx + ey + f\)$. The question involves a homogeneous quadratic part: $\(ax^2 + bxy + cy^2\)$. | $\(2x^2 + xy - 3y^2\)$ is an example. |
| Linear Factors | Expressions of the form $\(ax + by + c\)$. In homogeneous quadratics, the factors are typically of the form $\(ax + by\)$. | $\(2x + 3y\)$ and $\(x - y\)$ are linear factors in $\(x\)$ and $\(y\)$. |
| Factoring Quadratic Trinomials | Finding two binomials whose product is the given quadratic trinomial. For expressions like $\(ax^2 + bxy + cy^2\)$, we look for factors of the form $\((px + qy)(rx + sy)\)$. | Method often involves splitting the middle term or trial and error with coefficients. |
| Sum of Factors | Adding the resulting linear expressions together by combining like terms. | $\((ax + by) + (cx + dy) = (a+c)x + (b+d)y\)$ |
Additional Information: Homogeneous Quadratic Expressions
The expression $\(2x^2 + xy - 3y^2\)$ is an example of a homogeneous quadratic expression because all terms have the same degree (degree 2). Factoring such expressions often yields homogeneous linear factors (terms of degree 1).
Factoring homogeneous quadratics like $\(ax^2 + bxy + cy^2\)$ is very similar to factoring $\(at^2 + bt + c\)$ by substituting $\(t = x/y\)$ or by treating it as a quadratic in one variable (say, $\(x\)$, with $\(y\)$ as a constant) and then adjusting the terms.
For $\(2x^2 + xy - 3y^2\)$: If we treat it as a quadratic in $\(x\)$, it's $\(2x^2 + (y)x + (-3y^2)\)$. We look for factors $\((2x + Ay)(x + By)\)$. Expanding this gives $\(2x^2 + (2B+A)xy + Aby^2\)$. Comparing coefficients: Coefficient of $\(xy\)$: $\(2B + A = 1\)$ Coefficient of $\(y^2\)$: $\(AB = -3\)$ From $\(AB = -3\)$, possible integer pairs $\((A, B)\)$ are \((1, -3), (-1, 3), (3, -1), (-3, 1)\). Let's test these pairs in $\(2B + A = 1\)$: If $\((A, B) = (1, -3)\)$: $\(2(-3) + 1 = -6 + 1 = -5 \neq 1\)$. If $\((A, B) = (-1, 3)\)$: $\(2(3) + (-1) = 6 - 1 = 5 \neq 1\)$. If $\((A, B) = (3, -1)\)$: $\(2(-1) + 3 = -2 + 3 = 1\)$. This works! So $\(A=3, B=-1\)$. If $\((A, B) = (-3, 1)\)$: $\(2(1) + (-3) = 2 - 3 = -1 \neq 1\)$.
The working pair is $\(A=3\)$ and $\(B=-1\)$. Substituting these back into $\((2x + Ay)(x + By)\)$ gives $\((2x + 3y)(x - y)\)$. This confirms our previous factoring result.
The sum of the linear factors is indeed $\((2x + 3y) + (x - y) = 3x + 2y\)$.
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