Problem 29
Question
Factor completely. Identify any prime polynomials. $$ x^{2} y-13 x y+36 y $$
Step-by-Step Solution
Verified Answer
The completely factored form is: \( y (x - 9)(x - 4) \).
1Step 1 - Factor out the greatest common factor (GCF)
First, identify the greatest common factor in the expression. Here, the GCF is y, which is common in all the terms: \( x^{2} y - 13 x y + 36 y = y (x^2 - 13 x + 36) \)
2Step 2 - Factor the quadratic expression
Now, focus on the quadratic expression inside the parentheses: \( x^2 - 13 x + 36 \). We need to factor this quadratic expression by finding two numbers that multiply to 36 (the constant term) and add up to -13 (the coefficient of x). The numbers that fit are -9 and -4 because \( (-9) \times (-4) = 36 \) and \( (-9) + (-4) = -13 \). Therefore, the expression can be factored as: \( x^2 - 13 x + 36 = (x - 9)(x - 4) \)
3Step 3 - Combine the factored parts
Substitute the factored quadratic expression back into the overall expression: \( y (x - 9)(x - 4) \). This is the completely factored form of the original expression.
Key Concepts
greatest common factorquadratic expressionfactoring techniques
greatest common factor
Factoring out the greatest common factor (GCF) is the first step in simplifying expressions. The GCF is the highest number or variable that evenly divides each term in a polynomial. It helps in breaking down complex expressions into simpler factors. For the given example, the polynomial is: \[ x^{2} y - 13 x y + 36 y \]
The terms involve a common factor, which is 'y'.
Extracting y from each term makes it easier to handle:
\[ y (x^2 - 13 x + 36) \]
This reduction forms the base for further factoring steps.
Remembering to always start with the GCF can simplify the process significantly.
The terms involve a common factor, which is 'y'.
Extracting y from each term makes it easier to handle:
\[ y (x^2 - 13 x + 36) \]
This reduction forms the base for further factoring steps.
Remembering to always start with the GCF can simplify the process significantly.
quadratic expression
A quadratic expression is a polynomial of degree 2, generally in the form of: \[ ax^2 + bx + c \]
In our example, inside the parentheses, \[ x^2 - 13 x + 36 \]
We focus on factoring this quadratic.
This means we need to break it down into two binomials.
We can do this by finding two numbers that:
Here, the numbers are -9 and -4 because:
\[ (-9) \times (-4) = 36 \text{ and } (-9) + (-4) = -13 \]
This allows us to write the expression as:
\[ (x - 9)(x - 4) \]
This factorization technique helps break down quadratic expressions easily.
In our example, inside the parentheses, \[ x^2 - 13 x + 36 \]
We focus on factoring this quadratic.
This means we need to break it down into two binomials.
We can do this by finding two numbers that:
- Multiply to the constant term (36)
- Add up to the coefficient of the linear term (-13)
Here, the numbers are -9 and -4 because:
\[ (-9) \times (-4) = 36 \text{ and } (-9) + (-4) = -13 \]
This allows us to write the expression as:
\[ (x - 9)(x - 4) \]
This factorization technique helps break down quadratic expressions easily.
factoring techniques
Various factoring techniques can be used to simplify polynomials. Understanding these techniques is crucial for breaking down and solving complex expressions.
Let's review the key techniques:
Combining these techniques improves efficiency in simplifying polynomials.
For the provided example, we used the GCF to factor out 'y' and then factored the quadratic part to get the simplest form.
\[ y (x - 9)(x - 4) \]
Mastering these techniques will make polynomial factoring much more manageable and straightforward.
Let's review the key techniques:
- Greatest Common Factor (GCF): Extracting the highest shared factor across all terms.
- Factoring Quadratics: Splitting quadratic expressions into binomials by finding suitable pairs of numbers.
- Difference of Squares: Using the form \[ a^2 - b^2 = (a + b)(a - b) \]
- Perfect Square Trinomials: Recognizing patterns like \[ a^2 + 2ab + b^2 = (a + b)^2 \]
Combining these techniques improves efficiency in simplifying polynomials.
For the provided example, we used the GCF to factor out 'y' and then factored the quadratic part to get the simplest form.
\[ y (x - 9)(x - 4) \]
Mastering these techniques will make polynomial factoring much more manageable and straightforward.
Other exercises in this chapter
Problem 28
(a) factor out the greatest common factor. Identify any prime polynomials. (b) check. $$ 80 h^{2} k^{2}+24 h^{2} k+64 h k^{2}+240 $$
View solution Problem 29
A student said that the solutions of \((x+2)(x-6)=15\) are \(x=-2\) and \(x=6\), since \(-2+2=0\) and \(6-6=0\). Explain what is wrong with this thinking.
View solution Problem 29
Use a pattern to factor. Check. Identify any prime polynomials. $$ 100 f^{2}-9 h^{2} $$
View solution Problem 29
Use the \(a c\) method to factor. Check the factoring. Identify any prime polynomials. $$ 2 b^{2}-11 b-20 $$
View solution