Problem 105 - Beginning and Intermediate Algebra

Question

Find the following special products. Does \(4(t+3)^{2}=(4 t+12)^{2} ?\) Why or why not?

Step-by-Step Solution

Answer

No, \(4(t+3)^{2}\) is not equal to \((4t+12)^{2}\) because after expanding both sides, we get \(4t^2 + 24t + 36\) on the left side and \(16t^2 + 96t + 144\) on the right side, which are not equal expressions.

Step 1: Expand the left side of the equation

First, we need to expand (t+3)^2. Using the expansion formula (a+b)^2 = a^2 + 2ab + b^2, we get: \((t+3)^2 = t^2 + 2(t)(3) + 3^2\) This simplifies to: \((t+3)^2 = t^2 + 6t + 9\) Now, we need to multiply this expression by 4: \(4(t+3)^2 = 4(t^2 + 6t + 9)\)

Step 2: Apply distributive property to the left side

Now, we distribute the 4 to each term inside the parentheses: \(4(t^2 + 6t + 9) = 4t^2 + 24t + 36\)

Step 3: Expand the right side of the equation

Now we need to expand the right side of the equation, (4t+12)^2. Again, apply the expansion formula (a+b)^2 = a^2 + 2ab + b^2: \((4t+12)^2 = (4t)^2 + 2(4t)(12) + 12^2\) This simplifies to: \((4t+12)^2 = 16t^2 + 96t + 144\)

Step 4: Compare both sides of the equation

Now that we have found solutions for both sides of the equation, we need to compare them: Left side: \(4(t+3)^2 = 4t^2 + 24t + 36\) Right side: \((4t+12)^2 = 16t^2 + 96t + 144\) Since the left side is not equal to the right side, the equation is false.

Step 5: Conclusion

The expression \(4(t+3)^2\) is not equal to \((4t+12)^2\). Therefore, the equation is false.

Key Concepts

Polynomial Expansion

Polynomial expansion is a way to simplify expressions where you multiply polynomials. In the exercise, we expand \((t+3)^2\). Using the formula \((a+b)^2 = a^2 + 2ab + b^2\), you can break down the expression:

• First, square the first term \((a^2)\) which is \(t^2\).
• Then, calculate \(2ab\), which involves multiplying \(2\), \(t\), and \(3\), resulting in \(6t\).
• Finally, square the second term \((b^2)\) which is \(9\).

After adding them, you get \(t^2 + 6t + 9\). Polynomial expansion helps in simplifying and rearranging expressions to make computations easier. This is useful for finding products and powers of binomials.

Distributive Property

The distributive property is a key tool in algebra. It helps in expanding expressions by multiplying each term inside a parenthesis by a factor outside. For the problem, we apply it to \(4(t^2 + 6t + 9)\).
This involves:

• Multiplying \(4\) and \(t^2\) to get \(4t^2\).
• Then, multiplying \(4\) and \(6t\) to get \(24t\).
• Finally, multiplying \(4\) and \(9\) to get \(36\).

By distributing the \(4\) across all terms, you effectively simplify the equation. The distributive property makes handling larger polynomial expressions more manageable.

Quadratic Equations

Quadratic equations typically take the form \(ax^2 + bx + c = 0\). They can appear in many different contexts. In this exercise, although we're expanding rather than setting an equation to zero, understanding quadratics helps us see how expressions relate to quadratic forms.
They involve:

• Terms with squares \(t^2\), which help shape the curve of a parabola.
• Linear terms \(6t\) or \(24t\) that affect the slope.
• Constant terms \(9\) or \(36\) shift the position vertically.

Grasping concepts of quadratic equations aids in understanding the structure and behavior of polynomial expansions that resemble quadratic forms.

Binomial Theorem

The Binomial Theorem provides a powerful way to expand powers of binomials. While it isn't directly used in this exercise, understanding it can be very helpful. The theorem states:\[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]This formula allows you to expand any binomial raised to a power, with terms given by the coefficients \(\binom{n}{k}\), called binomial coefficients.

• For \((t+3)^2\), it simplifies easily using the formula, illustrating straightforward polynomial expansion.
• It supports understanding of how binomials behave algebraically when raised to powers.

Knowing the Binomial Theorem broadens your toolbox for tackling algebraic expressions efficiently.