Chapter 5: Problem 102
Factor. Assume that variables used as exponents represent positive integers. $$ x^{2 n}-7 x^{n}+12 $$
Short Answer
Expert verified
The factored form is \((x^n - 3)(x^n - 4)\).
Step by step solution
01
Identify the structure of the expression
The expression given is a quadratic in form: \[x^{2n} - 7x^n + 12\]. It can be rewritten using substitution: Let \( y = x^n \), then the expression becomes \[ y^2 - 7y + 12.\] This allows us to factor using familiar quadratic techniques.
02
Set up the quadratic equation for factoring
Given \( y = x^n \), the expression \( y^2 - 7y + 12 \) is a quadratic equation in standard form \( ay^2 + by + c \). In this case, \( a = 1 \), \( b = -7 \), and \( c = 12 \). We will factor to find two expressions that multiply to give the original quadratic.
03
Factor the quadratic using the product-sum method
With the quadratic \( y^2 - 7y + 12 \), find two numbers that multiply to \( 12 \) (the constant term) and add to \(-7\) (the coefficient of \( y \)). These numbers are \(-3\) and \(-4\), since \(-3 \times -4 = 12\) and \(-3 + (-4) = -7\). Thus, we can factor the quadratic as \((y - 3)(y - 4)\).
04
Substitute back the original variable
Replace \( y \) with \( x^n \) in the factored expression. Thus, the expression \((y - 3)(y - 4)\) becomes \((x^n - 3)(x^n - 4)\). This is the factored form of the original expression \(x^{2n} - 7x^n + 12\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique used to simplify algebraic expressions, especially when dealing with complex polynomials and, in our case, quadratic-like forms. It involves assigning a new variable to a part of the expression that looks familiar or simpler to solve.
Consider the expression given: \(x^{2n} - 7x^n + 12\). This resembles a quadratic format: something squared minus something linear plus a constant. Here, our "something" is \(x^n\), which we cleverly substitute with \(y\). This turns the expression into \(y^2 - 7y + 12\).
Using substitution, we transform challenging problems into more manageable ones. This method is particularly helpful:
Consider the expression given: \(x^{2n} - 7x^n + 12\). This resembles a quadratic format: something squared minus something linear plus a constant. Here, our "something" is \(x^n\), which we cleverly substitute with \(y\). This turns the expression into \(y^2 - 7y + 12\).
Using substitution, we transform challenging problems into more manageable ones. This method is particularly helpful:
- In polynomial equations where factorization isn't straightforward.
- When you spot repeating elements, as with \(x^n\).
- To transition into familiar mathematical territory, like simple quadratics.
Product-Sum Method
The product-sum method is a tried-and-tested technique used to factor quadratic equations. It is especially useful for quadratics expressed in standard form: \(ay^2 + by + c\). The key idea is to find two numbers that multiply to the product of \(a\) and \(c\), and simultaneously sum up to \(b\).
In our exercise, given \(y^2 - 7y + 12\) with \(a = 1\), \(b = -7\), and \(c = 12\):
In our exercise, given \(y^2 - 7y + 12\) with \(a = 1\), \(b = -7\), and \(c = 12\):
- We need two numbers whose product is \(12\) and sum is \(-7\).
- The numbers \(-3\) and \(-4\) meet these criteria: \(-3 \times -4 = 12\) and \(-3 + (-4) = -7\).
- Thus, we can directly factor our quadratic as \((y - 3)(y - 4)\).
Quadratic Equations
Quadratic equations are polynomial equations of degree two, typically in the form \(ax^2 + bx + c = 0\). They are fundamental in algebra, appearing in various mathematical fields. The exercise showcases a quadratic equation in disguise, \(x^{2n} - 7x^n + 12\).
Using substitution, we translated this expression into a clear quadratic \(y^2 - 7y + 12\), enhancing its manageability. Quadratics can be solved using numerous strategies:
Using substitution, we translated this expression into a clear quadratic \(y^2 - 7y + 12\), enhancing its manageability. Quadratics can be solved using numerous strategies:
- Factoring, as we did using the product-sum method.
- The quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
- Completing the square.