Problem 67 - Elementary and Intermediate Algebra

Question

Factor. If an expression is prime, so indicate. $$ 8 y^{2}-2 y-1 $$

Step-by-Step Solution

Answer

The expression factors to \((4y - 2)(2y + 1)\).

Step 1: Identify the Quadratic Expression

The given expression is a quadratic in the form of \( 8y^2 - 2y - 1 \). Our task is to factor it, if possible.

Step 2: Find Suitable Coefficients for Factoring

We need two numbers that multiply to \( 8 \times -1 = -8 \) and add to \(-2\). The possible pairs are \((-4, 2)\) and \((4, -2)\), but neither of these add to \(-2\), indicating that this expression is challenging to factor by conventional methods.

Step 3: Use the Quadratic Formula for Further Confirmation

To confirm if the expression is prime or not, we use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 8, \ b = -2, \ c = -1 \). Calculating the discriminant: \((b^2 - 4ac) = (-2)^2 - 4 \times 8 \times (-1) = 4 + 32 = 36\), which is a perfect square.

Step 4: Calculate the Roots Using Quadratic Formula

Substitute the values into the formula: \( y = \frac{2 \pm \sqrt{36}}{16} \). Hence, the roots are \( y = \frac{2 + 6}{16} = \frac{1}{2} \) and \( y = \frac{2 - 6}{16} = -\frac{1}{4} \).

Step 5: Write the Factored Form

Now that we know the roots, the quadratic can be expressed in factored form using the roots: \( 8(y - \frac{1}{2})(y + \frac{1}{4}) \). Simplifying, we multiply through by 4: \((4y - 2)(2y + 1)\).

Step 6: Check the Factored Form

Expand \((4y - 2)(2y + 1)\) to verify it matches the original expression: \( (4y)(2y) + (4y)(1) + (-2)(2y) + (-2)(1) = 8y^2 + 4y - 4y - 2 = 8y^2 - 2y - 2 \).

Key Concepts

Quadratic Formula

To solve quadratic equations, the quadratic formula is a reliable tool. It helps find the roots of any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula is given by:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula arises from completing the square on a general quadratic equation. Here, \( a \), \( b \), and \( c \) represent the coefficients of the terms in the equation:

• \( a \) is the coefficient of \( x^2 \)
• \( b \) is the coefficient of \( x \)
• \( c \) is the constant term

Using this formula, you can easily find both roots of the equation, provided the discriminant is non-negative. This can be quite handy if factoring the quadratic seems complex or impractical.

Roots of Quadratic Equation

Roots of a quadratic equation are the solutions or the values of \(x\) which satisfy the equation \(ax^2 + bx + c = 0\). These roots can be real or complex numbers depending on the discriminant value. Using the quadratic formula, the roots can be found as follows:

• \( x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \)
• \( x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \)

In the context of our quadratic \( 8y^2 - 2y - 1 = 0 \):

• Calculate \( b^2 - 4ac = (-2)^2 - 4 \times 8 \times (-1) = 36 \)
• Plug these into the formula to find \(y = \frac{2 \pm 6}{16}\)

This results in roots \( y = \frac{1}{2} \) and \( y = -\frac{1}{4} \). These roots indicate where the parabola defined by the quadratic intersects the x-axis.

Discriminant Calculation

The discriminant is crucial in determining the nature of the roots of a quadratic equation. It is found as part of the quadratic formula:

• The expression inside the square root, \( b^2 - 4ac \), is the discriminant.

The value of the discriminant reveals the type of roots:

• If \( b^2 - 4ac > 0 \), there are two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is one real repeated root.
• If \( b^2 - 4ac < 0 \), the roots are complex and not real.

For the quadratic \( 8y^2 - 2y - 1 \), substituting the coefficients into the discriminant formula gives \( 36 \). This value being positive and a perfect square indicates that the roots are real and rational, matching our calculated roots \( y = \frac{1}{2} \) and \( y = -\frac{1}{4} \). Understanding the discriminant helps decide the best approach to solving the quadratic equation.