Problem 86 - Algebra and Trigonometry Real Mathematics, Real People

Question

Factor the expression. $$4 y^{2}-28 y+49$$

Step-by-Step Solution

Answer

The factored form of the expression \(4y^{2} - 28y + 49\) is \((2y-7)^{2}\)

Step 1: Identifying the Quadratic Structure

The given expression \(4y^{2} - 28y + 49\) exhibits the structure of a quadratic expression, more specifically, it looks like a perfect square trinomial as \(a^{2} - 2ab + b^{2} = (a - b)^{2}\).

Step 2: Calculating Square Roots

In order to express the equation as a square of a binomial, calculate the square root of the first and last term. The square root of \(4y^{2}\) is \(2y\) and the square root of \(49\) is \(7\). The middle term is negative which indicates that the binomial will be in the form of \(a-b\). Now, since \(2*2y*7 = 28y\), it can be said that it's actually a perfect square trinomial.

Step 3: Writing in Factored Form

The quadratic expression can hence be written in factored form. So, it turns out to be \((2y-7)^{2}\). This is the factored form of the given quadratic expression.

Step 4: Verification

Verify the factored form by expanding \((2y-7)^{2}\). This gives us back the original quadratic expression \(4y^{2} - 28y + 49\), confirming that the factored form is correct.

Key Concepts

Perfect Square Trinomial

A perfect square trinomial is a special kind of quadratic expression that can be written as the square of a binomial. This means it takes the form \(a^2 - 2ab + b^2 = (a-b)^2\) or \(a^2 + 2ab + b^2 = (a+b)^2\). Recognizing these structures in expressions can make factoring much easier.

To determine if a trinomial is a perfect square, check the first and last terms to ensure they are perfect squares themselves. Then, see if the middle term fits the pattern \(2ab\) when considering the whole expression. In practice, for the expression \(4y^2 - 28y + 49\):

• Take the square root of \(4y^2\), which is \(2y\).
• Take the square root of \(49\), which is \(7\).
• Confirm that twice the product of these square roots, \(2*2y*7\), equals the middle term \(-28y\).

Since this holds true, \(4y^2 - 28y + 49\) is indeed a perfect square trinomial and can be factored as \((2y-7)^2\). Understanding this pattern helps simplify solving quadratic expressions by recognizing this structure.

Binomial Expansion

When dealing with binomials, expansion refers to the process of multiplying out the terms of a binomial raised to a power. For our expression \((2y-7)^2\), expanding means calculating \((2y-7) * (2y-7)\). This process confirms the original form of the quadratic expression.

The steps to expand this particular binomial are straightforward:

• Multiply the first terms: \((2y)*(2y) = 4y^2\).
• Multiply the inner terms: \((2y)*(-7) = -14y\).
• Multiply the outer terms: \((-7)*(2y) = -14y\).
• Multiply the last terms: \((-7)*(-7) = 49\).

Combine these results and the expression becomes \(4y^2 - 14y - 14y + 49\), which simplifies to \(4y^2 - 28y + 49\).

This methodical approach of breaking down the binomial multiplication ensures the factors correctly reconstruct the original expression, demonstrating the utility of the perfect square trinomial identification.

Quadratic Expressions

Quadratic expressions are polynomial expressions that involve terms up to the second degree. A standard quadratic expression in one variable is written as \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants.

Understanding the types of quadratic expressions is essential in solving and simplifying them. There are a few different forms and ways they can be manipulated:

• Standard form: \(ax^2 + bx + c\)
• Factored form: For instance, \((x-p)(x-q)\) which expands to \(x^2 - (p+q)x + pq\).
• Vertex form: \(a(x-h)^2 + k\), indicating a parabola with its vertex at \((h, k)\).

In working with quadratic expressions, one might aim to transform them into a factored form for solving quadratic equations or rewriting them as vertex form to understand their graph better.

In the provided step-by-step solution, the expression \(4y^2 - 28y + 49\) is expertly identified as a perfect square trinomial. This conclusion, resulting in the factorization into \((2y-7)^2\), highlights the power of recognizing underlying patterns in quadratics to simplify otherwise complex expressions.