Question

use distribution to find the value of (a+b)(a+b) with no parentheses.
pythagorean theorem
a = a1 + a2 + a3 + a4 + a5
(a+b)² = a1 + a2 + a3 + a4 + a5
(a+b)² = a1 + a2 + a3 + a4 + c²
(a+b)² = ½ab + ½ab ½ab + ½ab + c²
(a+b)(a+b) = 2ab + c²
options:
$a^2 + 2ab + b^2$
$a^2 + b^2$
$2a + 2ab + 2b$
$2(a + b)$

Explanation:

Step1: Apply Distributive Property

We use the distributive property (also known as the FOIL method for binomials) to expand \((a + b)(a + b)\). The distributive property states that for any terms \(x\), \(y\), and \(z\), \(x(y + z)=xy+xz\). So, we treat the first \((a + b)\) as a single term and distribute it over the second \((a + b)\):

$$ (a + b)(a + b)=a(a + b)+b(a + b) $$

Step2: Distribute Again

Now, we distribute \(a\) over \((a + b)\) and \(b\) over \((a + b)\) separately:

$$ a(a + b)=a\times a+a\times b=a^{2}+ab $$

$$ b(a + b)=b\times a + b\times b=ab + b^{2} $$

Step3: Combine Like Terms

We substitute these back into the expression from Step 1:

$$ (a + b)(a + b)=(a^{2}+ab)+(ab + b^{2}) $$

Now, we combine the like terms \(ab\) and \(ab\):

$$ (a + b)(a + b)=a^{2}+(ab + ab)+b^{2}=a^{2}+2ab + b^{2} $$

We can also verify this using the area model provided. The large square has side length \((a + b)\), so its area is \((a + b)^{2}\). The area can also be calculated as the sum of the areas of the two smaller squares (\(a^{2}\) and \(b^{2}\)) and the two rectangles (each with area \(ab\), so total \(2ab\)). Thus, the area is \(a^{2}+2ab + b^{2}\), which matches our algebraic expansion.

Answer:

\(a^{2}+2ab + b^{2}\) (corresponding to the option \(a^{2}+2ab + b^{2}\))