Problem 65
Question
Factor $$ x^{2}+x-30 $$
Step-by-Step Solution
Verified Answer
The expression \(x^2 + x - 30\) factors to \((x + 6)(x - 5)\).
1Step 1: Identify the quadratic expression
The expression given for factoring is \(x^2 + x - 30\). This is a quadratic expression of the form \(ax^2 + bx + c\), where \(a = 1\), \(b = 1\), and \(c = -30\).
2Step 2: Find two numbers that multiply to \(c\)
We need to find two numbers that multiply to \(c = -30\) and add up to \(b = 1\). These numbers are \(6\) and \(-5\) because \(6 imes (-5) = -30\) and \(6 + (-5) = 1\).
3Step 3: Write the middle term as a sum of two terms based on step 2
Rewrite the quadratic expression by splitting the middle term \(x\) into two terms using the numbers from the previous step. The expression becomes \(x^2 + 6x - 5x - 30\).
4Step 4: Factor by grouping
Group the first two terms and the last two terms: \((x^2 + 6x) + (-5x - 30)\). Factor each group separately: \(x(x + 6) - 5(x + 6)\).
5Step 5: Factor out the common binomial factor
Since both terms contain \((x + 6)\), factor \((x + 6)\) out: \((x + 6)(x - 5)\). Thus, the factored form of \(x^2 + x - 30\) is \((x + 6)(x - 5)\).
Key Concepts
Quadratic ExpressionFactor by GroupingBinomial Factor
Quadratic Expression
A quadratic expression is a polynomial of degree two, which means the highest power of the variable is squared (usually represented as \( x^2 \)).
This form is generally written as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants.
The term "quadratic" comes from "quadratus" in Latin, meaning square.To identify a quadratic expression, take a look at the highest power of the variable.
If it's two, then you’re dealing with a quadratic expression.
In our example, \( x^2 + x - 30 \), we identify it by noting the initial \( x^2 \) term.Here, the coefficient \( a \) is 1, \( b \) is 1, and \( c \) is \(-30\).
Understanding these coefficients helps in the factoring process, particularly in deciding how to split the middle term.
This form is generally written as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants.
The term "quadratic" comes from "quadratus" in Latin, meaning square.To identify a quadratic expression, take a look at the highest power of the variable.
If it's two, then you’re dealing with a quadratic expression.
In our example, \( x^2 + x - 30 \), we identify it by noting the initial \( x^2 \) term.Here, the coefficient \( a \) is 1, \( b \) is 1, and \( c \) is \(-30\).
Understanding these coefficients helps in the factoring process, particularly in deciding how to split the middle term.
Factor by Grouping
Factor by grouping is a useful method for factoring certain types of polynomial expressions.
It works particularly well for expressions where grouping terms allows for the extraction of a common binomial factor.In this method:
Factor each group separately, i.e., \( x(x + 6) \) and \(-5(x + 6)\).
Finally, you can factor out the common binomial \((x + 6)\).
It works particularly well for expressions where grouping terms allows for the extraction of a common binomial factor.In this method:
- You first separate the expression into two groups.
- Next, factor out the greatest common factor from each group.
- After that, look for a common binomial in both parts that can be factored out.
Factor each group separately, i.e., \( x(x + 6) \) and \(-5(x + 6)\).
Finally, you can factor out the common binomial \((x + 6)\).
Binomial Factor
A binomial factor is a polynomial with just two terms.
This factor is what’s often extracted last when you factor by grouping.Once you've separated your expression by grouping and factored each group, you should observe a common binomial factor.
For the expression \( x^2 + x - 30 \), after splitting terms and grouping effectively, we identified \((x + 6)\) as common binomial factor in the grouped terms.Extracting this common binomial leads to the completely factored form. In our example, once the \((x + 6)\) is factored out, we are left with \( (x + 6)(x - 5) \).
This step is crucial because it simplifies the quadratic expression into a product of two binomial factors, revealing its simpler "roots" or solutions.
This factor is what’s often extracted last when you factor by grouping.Once you've separated your expression by grouping and factored each group, you should observe a common binomial factor.
For the expression \( x^2 + x - 30 \), after splitting terms and grouping effectively, we identified \((x + 6)\) as common binomial factor in the grouped terms.Extracting this common binomial leads to the completely factored form. In our example, once the \((x + 6)\) is factored out, we are left with \( (x + 6)(x - 5) \).
This step is crucial because it simplifies the quadratic expression into a product of two binomial factors, revealing its simpler "roots" or solutions.
Other exercises in this chapter
Problem 65
Find \(h(5)\) and \(h(-2) .\) See Example 4. $$ h(x)=\frac{x^{2}+2 x-35}{x^{2}+5 x+6} $$
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Factor each expression completely. Factor a difference of two squares first. \(x^{6}-1\)
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Solve each compound inequality, if possible. Graph the solution set (if one exists) and write it using interval notation. $$ -24
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Solve each inequality. Graph the solution set and write it using interval notation. \(|x-12|>24\)
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