Expand (2x + 5)²
Watch the identity (a+b)² = a² + 2ab + b² in action with real coefficients.
Equation
y = (2*x + 5)^2 - (4*x^2 + 0*x*5 + 5^2)
Graph
Table
| x | y |
|---|---|
| -5 | -100 |
| -4 | -80 |
| -3 | -60 |
| -2 | -40 |
| -1 | -20 |
| 0 | 0 |
| 1 | 20 |
| 2 | 40 |
| 3 | 60 |
| 4 | 80 |
| 5 | 100 |
What this lesson covers
What you do
You shape the function y = (2*x + a)^2 - (4*x^2 + k*x*a + a^2) and watch the graph answer.
Challenges to clear
- (2x+a)²: the middle term is 2·(2x)·a = 4xa. Slide k until the difference is 0 at x = 2.
- 0 at x = −1 too — with real coefficients the identity still holds.
- Same idea, a steeper binomial: (3x + a)². Set a = 5.
- The middle term is 2·(3x)·a = 6xa. Slide k until the difference is 0 at x = 2.
Check yourself
Why does the graph staying flat at y=0 prove the identity?
What is the common mistake when expanding (2x + a)²?
If a=3, what is the expanded form of (2x + 3)²?
- Because the graph is a straight line.
- Because LHS - RHS = 0 means LHS = RHS for all x and a. — correct
- Because the parameter 'a' is always positive.
- Because x is squared in both terms.
- Forgetting to square the 2x term.
- Writing (2x)² + a² and forgetting the middle term 2*(2x)*a. — correct
- Adding the terms instead of multiplying.
- Using the wrong sign for the middle term.
- 4x² + 9
- 4x² + 6x + 9
- 4x² + 12x + 9 — correct
- 2x² + 12x + 9
Think about it
- (2x+5)²: the middle term is 2 · (2x) · 5 = 20x. With a = 5, what k gives k·x·a = 20x?