Solve : (x + 2y) (2x – y) A. 2x 2+ 5xy – 2y 2 B. 2x 2+ 3xy – 2y 2 C. x 2+ 4xy + y 2 D. x 2+ 4xy – y 2
B
Solving Polynomial Multiplication: \((x + 2y)(2x - y)\)
The problem asks us to find the product of two binomials: \((x + 2y)\) and \((2x - y)\). We can solve this by using the distributive property, often remembered by the acronym FOIL, which stands for First, Outer, Inner, Last.
Let's break down the multiplication step-by-step:
- First: Multiply the first terms in each binomial.
- Outer: Multiply the outer terms of the binomials.
- Inner: Multiply the inner terms of the binomials.
- Last: Multiply the last terms in each binomial.
Now, let's apply these steps to the given expression \((x + 2y)(2x - y)\):
- First: Multiply \(x\) and \(2x\).
\(x \times (2x) = 2x^2\) - Outer: Multiply \(x\) and \(-y\).
\(x \times (-y) = -xy\) - Inner: Multiply \(2y\) and \(2x\).
\((2y) \times (2x) = 4xy\) - Last: Multiply \(2y\) and \(-y\).
\((2y) \times (-y) = -2y^2\)
Now, we combine these four results:
\[2x^2 + (-xy) + 4xy + (-2y^2)\]
\[2x^2 - xy + 4xy - 2y^2\]
Next, combine the like terms. The like terms here are \(-xy\) and \(4xy\):
\[-xy + 4xy = (-1 + 4)xy = 3xy\]
Substitute this back into the expression:
\[2x^2 + 3xy - 2y^2\]
This is the simplified product of \((x + 2y)(2x - y)\).
Let's compare our result to the given options:
| Option | Expression |
|---|---|
| A | \(2x^2 + 5xy - 2y^2\) |
| B | \(2x^2 + 3xy - 2y^2\) |
| C | \(x^2 + 4xy + y^2\) |
| D | \(x^2 + 4xy - y^2\) |
Our calculated result, \(2x^2 + 3xy - 2y^2\), matches Option B.
Revision Table: Key Concepts in Polynomial Multiplication
| Concept | Description | Example |
|---|---|---|
| Binomial | A polynomial with two terms. | \(x+2y\), \(2x-y\) |
| Term | A single number, variable, or product of numbers and variables. | \(x\), \(2y\), \(2x\), \(-y\) |
| Like Terms | Terms that have the same variables raised to the same powers. | \(-xy\) and \(4xy\); \(3a^2b\) and \(-5a^2b\) |
| Distributive Property | Multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the products. For binomials, this means each term in the first binomial is multiplied by each term in the second binomial. | \(a(b+c) = ab + ac\) \((a+b)(c+d) = a(c+d) + b(c+d) = ac + ad + bc + bd\) |
| FOIL Method | A mnemonic for multiplying two binomials: First, Outer, Inner, Last. It ensures all terms are multiplied. | Explained in the solution steps above. |
Additional Information: Expanding Polynomials
Expanding polynomials involves multiplying them together to remove parentheses. The distributive property is the fundamental principle used. For polynomials with more than two terms (like a binomial multiplied by a trinomial), you still multiply each term in the first polynomial by every term in the second polynomial.
For example, to multiply \((x+y)(x^2+2xy+y^2)\):
- Multiply \(x\) by each term in the second polynomial: \(x(x^2+2xy+y^2) = x^3 + 2x^2y + xy^2\)
- Multiply \(y\) by each term in the second polynomial: \(y(x^2+2xy+y^2) = x^2y + 2xy^2 + y^3\)
- Combine the results: \((x^3 + 2x^2y + xy^2) + (x^2y + 2xy^2 + y^3) = x^3 + (2x^2y + x^2y) + (xy^2 + 2xy^2) + y^3\)
- Combine like terms: \(x^3 + 3x^2y + 3xy^2 + y^3\)
Polynomial multiplication is a core skill in algebra used in various contexts, including solving equations, graphing functions, and calculus.
Important Questions from Quadratic Equation
If the equations x 2+ ax + b = 0 and x 2+ bx + a = 0 have a common root, then find the value of a + b (where a is not equal to b)
If x 2+ 1 = 2x, then find x – \((\frac{1}{x})\)
- Roots of the following equation are 6x 2+ 4x - 2 = 0
Find the factors of (x 2– x – 132)?
Which of the following is NOT a quadratic equation?
