Problem 58
Question
Factor completely. $$5 g^{2}+g-7$$
Step-by-Step Solution
Verified Answer
The completely factored expression of \(5g^2 + g - 7\) is \((5g + 7)(g - 1)\).
1Step 1: Identify the AC method components
In the given expression, \(5g^2 + g - 7\), we have \(a=5\), \(b=1\), and \(c=-7\). The AC method states that we should multiply \(a\) and \(c\), so we get: \(ac = 5 \times -7 = -35\).
2Step 2: Find two numbers that multiply to \(ac\) and add to \(b\)
We need to find two numbers, \(m\) and \(n\), such that \(mn = ac = -35\) and \(m + n = b = 1\). After considering various factors of -35, we find that \(m = 7\) and \(n = -5\) satisfy these conditions, since \(7 \times -5 = -35\) and \(7 - 5 = 1\).
3Step 3: Rewrite the expression with the new terms
Replace the middle term (\(g\)) with the two new terms found in the previous step (\(7g - 5g\)):
\(5g^2 + 7g - 5g -7\).
4Step 4: Factor by grouping
In this step, we'll factor the expression in pairs. Group the first two terms and the last two terms:
\((5g^2 + 7g) - (5g + 7)\).
Now, factor out the greatest common factor (GCF) from each group:
\(g(5g + 7) - 1(5g + 7)\).
5Step 5: Factor out the common binomial
We can now see that both terms have a common binomial factor \((5g + 7)\). Factor this out of each term:
\((5g + 7)(g - 1)\).
So, the completely factored expression for the given quadratic is \((5g + 7)(g - 1)\).
Key Concepts
AC methodPolynomial expressionsFactor by grouping
AC method
The AC method is a fantastic technique for factoring quadratic equations, especially when the quadratic cannot be easily factored by inspection. The name 'AC method' comes from the idea of multiplying the coefficient of the quadratic term (\(a\)) with the constant term (\(c\)). With this approach, we transform the trinomial into a simpler form that can be factored by grouping.
For example, for the polynomial \(5g^2 + g - 7\), the values are \(a = 5\), \(b = 1\), and \(c = -7\).
For example, for the polynomial \(5g^2 + g - 7\), the values are \(a = 5\), \(b = 1\), and \(c = -7\).
- First, multiply the \(a\) term by the \(c\) term: \(ac = 5 \times -7 = -35\).
- Then, find two numbers that both multiply to \(-35\) and add to \(1\) (the \(b\) coefficient).
- Through trial and exploration, identify \(7\) and \(-5\) as these numbers, because \(7 \cdot -5 = -35\) and \(7 + -5 = 1\).
Polynomial expressions
When working with polynomial expressions, you need to understand the structure and degree of the polynomial. A polynomial is a mathematical expression that consists of variables and coefficients, involving operations of addition, subtraction, multiplication, and non-negative integer exponents.
Quadratic polynomials, like \(5g^2 + g - 7\), are expressions of the form \(ax^2 + bx + c\). Here, the variable is \(g\), and it has a degree of 2, meaning the highest exponent on the variable is 2.
Quadratic polynomials, like \(5g^2 + g - 7\), are expressions of the form \(ax^2 + bx + c\). Here, the variable is \(g\), and it has a degree of 2, meaning the highest exponent on the variable is 2.
- The coefficients \(a\), \(b\), and \(c\) correspond to the terms in order of their power.
- If you are given a quadratic polynomial, your aim is often to factor it down into a product of simpler binomials, like turning \(5g^2 + g - 7\) into \((5g + 7)(g - 1)\).
- Factoring helps in simplifying expressions and solving equations easily as it reveals the roots or solutions of the polynomial equation when set to zero.
Factor by grouping
Factor by grouping is an effective method for factoring polynomials, particularly when it comes to quadratics structured in a way where two pairs of terms can be grouped. This approach involves grouping terms in pairs and factoring out the greatest common factor from each pair.
To illustrate, use the expression \(5g^2 + 7g - 5g - 7\):
To illustrate, use the expression \(5g^2 + 7g - 5g - 7\):
- First, split the expression into two groups: \((5g^2 + 7g) - (5g + 7)\).
- Next, factor out the greatest common factor from each of these groups: \(g(5g + 7)\) and \(-1(5g + 7)\).
- Once you factor each group, what's left is a common binomial factor, \((5g + 7)\).
- Finally, factor out this common factor to get \((5g + 7)(g - 1)\).
Other exercises in this chapter
Problem 58
Solve each equation. $$(6 n+5)^{2}-(3 n+4)^{2}=0$$
View solution Problem 58
Factor by grouping. \(3 j k-7 k+6 j-14\)
View solution Problem 58
Factor completely, if possible. Begin by asking yourself, "Can I factor out a GCF?" $$6 g^{5}+6 g^{4}-12 g^{3}$$
View solution Problem 59
Factor completely. $$b^{4}-64$$
View solution