Question

Factor each polynomial completely. If the polynomial cannot be factored, say it is prime. $$ 9 y^{2}+9 y-4 $$

Short answer

(3y - 1)(3y + 4)

Step-by-step solution

Identify the polynomial structure

The given polynomial is a quadratic expression of the form: ax^2 + bx + cwhere a = 9, b = 9, and c = -4.

Calculate the Discriminant

The discriminant of a quadratic equation x^2 + bx + c is given by the formula d = b^2 - 4acWe substitute a, b, and c: d = 9^2 - 4(9)(-4) = 81 + 144 = 225Since the discriminant is a perfect square (225), the polynomial has rational roots and can be factored.

Find the roots using the Quadratic formula

Quadratic roots are given by: \[ x = \frac{-b \, \pm \, \sqrt{d}}{2a} \] Substitute the values a = 9, b = 9, and d = 225: \[ x = \frac{-9 \, \pm \, 15}{2(9)} \] This gives us: \[ x_1 = \frac{-9 + 15}{18} = \frac{6}{18} = \frac{1}{3} \] \[ x_2 = \frac{-9 - 15}{18} = \frac{-24}{18} = -\frac{4}{3} \]

Write expression in factorized form

Using the roots found, the polynomial can be factored as: \[ (3y - 1)(3y + 4) \]

Key concepts

quadratic equations

A quadratic equation is a type of polynomial that has a degree of 2. This means it includes a term with the variable raised to the power of 2. The general form of a quadratic equation is \( ax^2 + bx + c \). Here, 'a', 'b', and 'c' are constants, with 'a' not equal to zero.
Quadratic equations can be solved through various methods, including factoring, completing the square, and using the quadratic formula. Quadratic equations are core to many algebra problems and appear in numerous real-world scenarios, such as physics and engineering.
Recognizing a quadratic equation quickly and knowing the suitable method to solve it is essential. Always check first if it can be simplified or factored before diving into more complex methods.

discriminant

The discriminant is a key concept in solving quadratic equations. It is a part of the quadratic formula, \( x = \frac{-b \, \text{±} \, \text{√d}}{2a} \), where \( d = b^2 - 4ac \). The discriminant can tell us important information about the nature of the roots without solving the equation.

• If the discriminant is positive and a perfect square, the quadratic equation has two distinct rational roots.
• If it is positive but not a perfect square, the roots are distinct and irrational.
• If the discriminant is zero, there is exactly one real root, meaning the quadratic touches the x-axis at one point.
• If the discriminant is negative, the quadratic equation will have two complex (imaginary) roots.

In the given exercise, the discriminant calculates to 225, which is a perfect square. This indicates the quadratic equation has two distinct rational roots.

factoring techniques

Factoring a quadratic equation means expressing it as a product of two binomials.
There are several techniques used to factor quadratics:

• **Factoring by grouping:** Often used when the quadratic is not easily factorable.
• **Using special products:** Recognizing patterns such as perfect square trinomials and the difference of squares.
• **AC method:** Useful when 'a' (the coefficient of \(x^2\)) is greater than 1.
• **Quadratic formula and roots:** Calculates the roots first and then derives the factored form from them, as shown in the solution.

Once the roots are identified, the original quadratic can be written as \(a(x - r_1)(x - r_2) \). In the exercise provided, after finding the roots, 1/3 and -4/3 must be rearranged in the form \((3y-1)(3y+4)\).

roots of quadratic equations

The roots (or solutions) of quadratic equations are the values of the variable that make the equation true. In other words, they are the points where the graph of the quadratic equation intersects the x-axis.
There are different methods to find the roots:

• The quadratic formula: \( x = \frac{-b \, \text{±} \, \text{√(b^2 - 4ac)}}{2a} \) is a universal method that works for all quadratic equations.
• Factoring: Expressing the quadratic in the product form \((x - r_1)(x - r_2)\) directly gives the roots \(r_1\) and \(r_2\).
• Graphing: Visually inspecting where the graph crosses the x-axis.

For the provided polynomial \(9 y^2 + 9 y - 4\), the roots were found using the quadratic formula as \( y = \frac{1}{3} \) and \( y = -\frac{4}{3} \). Converting them to factor form results in the expression \((3y - 1)(3y + 4)\).