Problem 30 - Algebra 1: Concepts and Skills

Question

Factor the trinomial. $$ 12 y^{2}-20 y+8 $$

Step-by-Step Solution

Answer

The factorized form of the trinomial is \( 4 * (3y - 2)(y - 1) \)

Step 1: STEP 1: Factoring out the common factor

First, checking the three terms of the trinomial, a common factor can be found, which is 4. This is then factored out to get a simpler trinomial: \( 12y^{2} - 20y + 8 = 4(3y^{2} - 5y + 2) \)

Step 2: STEP 2: Factorizing the simpler trinomial

Now the problem comes down to factorizing the simpler trinomial \( 3y^{2} - 5y + 2 \). To do this, look for two numbers that multiply to give the product of the coefficient of \( y^{2} \) (i.e., 3) and the constant term (i.e., 2), which is 6; and also add to give the coefficient of \( y \) (i.e., -5). The numbers -2 and -3 satisfy these conditions. Hence we can write the factors as:\[ 3y^{2} - 5y + 2 = (3y - 2)(y - 1) \]

Step 3: STEP 3: Writing the final factorized form

The final answer, then, is gotten by factoring out the common factor (4) we discovered in STEP 1 with the factored form of the trinomial in STEP 2:\[ 12y^{2} - 20y + 8 = 4 * (3y - 2)(y - 1) \]

Key Concepts

Greatest Common Factor

When faced with a polynomial, finding the greatest common factor (GCF) is a smart first step. The GCF is the largest number or expression that divides all the terms of the polynomial without leaving a remainder. In the exercise, the trinomial \(12y^2 - 20y + 8\) has three terms. We observe that each term has the number 4 as a factor. Thus, 4 is the GCF.To factor out the GCF:

• Divide every term in the trinomial by 4.
• Represent the trinomial as a product of 4 and another trinomial: \(4(3y^2 - 5y + 2)\).

By factoring out the GCF, you simplify the expression, making it easier to handle.

Trinomial Factoring

Trinomial factoring involves breaking down a polynomial in the form of \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. The goal is to express it as a product of two binomials.For the trinomial \(3y^2 - 5y + 2\), the aim is to find two numbers that:

• Multiply to \(a \times c\), which is \(3 \times 2 = 6\).
• Add up to \(b\), which is \(-5\).

The numbers \(-2\) and \(-3\) work because \(-2 \times -3 = 6\) and \(-2 + -3 = -5\). Thus, the trinomial can be rewritten by splitting the middle term:\[3y^2 - 5y + 2 = 3y^2 - 3y - 2y + 2\]Next, group terms and factor by grouping:

• \((3y^2 - 3y) + (-2y + 2)\)
• Factor each group: \(3y(y - 1) - 2(y - 1)\)

Finally, combine into binomials:\[(3y - 2)(y - 1)\]

Quadratic Expressions

Quadratic expressions, like \(ax^2 + bx + c\), are a specific type of polynomial where the highest exponent is 2. The roots or solutions of these expressions represent the values for which the expression equals zero.In this exercise, we started with the quadratic expression \(12y^2 - 20y + 8\). After factoring out the GCF, we simplified it to \(3y^2 - 5y + 2\) and from there factored it further into binomials.Why Factor Quadratics?

• Solving quadratics: Factoring allows for a straightforward method to find the roots.
• Simplifying expressions: You reduce complexity by breaking down expressions.
• Applications: Widely used in physics, engineering, and economics to model scenarios.

Understanding each step in factoring helps uncover how quadratics behave and provides solutions quickly.