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Magazine Chapter 4: Problem 11
Factor completely. \(x^{4}-23 x^{2}-50\)
Short Answer
Expert verified
Factored form: (x-5)(x+5)(x^2+2).
Step by step solution
01
Identify Quadratic Form
The expression \(x^4 - 23x^2 - 50\) can be treated like a quadratic in form by recognizing \(x^4\) as \((x^2)^2\) and \(x^2\) as \(u\). Thus, rewrite the expression as \(u^2 - 23u - 50\).
02
Use the Quadratic Formula
To factor \(u^2 - 23u - 50\), consider using the quadratic formula, which is given by \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -23\), and \(c = -50\).
03
Calculate the Discriminant
Calculate the discriminant \(b^2 - 4ac\). We find \((-23)^2 - 4 \times 1 \times (-50) = 529 + 200 = 729\).
04
Solve for the Roots
With the discriminant \(729\), solve for \(u\) using the quadratic formula: \(u = \frac{23 \pm \sqrt{729}}{2} = \frac{23 \pm 27}{2}\).
05
Identify and Substitute Roots
The roots are \(u = 25\) and \(u = -2\). Substitute \(u = x^2\) back into these solutions: \(x^2 = 25\) and \(x^2 = -2\).
06
Factor the Original Expression
Determine the factors. \(x^2 - 25\) factors further into \((x-5)(x+5)\), since 25 is a perfect square. \(x^2 + 2\) does not factor further over the real numbers.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a powerful tool used to find the roots of a quadratic equation of the form\( ax^2 + bx + c = 0 \). It is expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a \), \( b \), and \( c \) are known values from the quadratic equation.
In our exercise, we used a substitution \( u = x^2 \) to transform the polynomial into a quadratic form \( u^2 - 23u - 50 \). This allowed us to apply the quadratic formula to find the values of \( u \).
- \( a \) is the coefficient of \( x^2 \)
- \( b \) is the coefficient of \( x \)
- \( c \) is the constant term
In our exercise, we used a substitution \( u = x^2 \) to transform the polynomial into a quadratic form \( u^2 - 23u - 50 \). This allowed us to apply the quadratic formula to find the values of \( u \).
Discriminant
The discriminant is a part of the quadratic formula that lies under the square root symbol: \( b^2 - 4ac \). This value can give insight into the nature of the roots of the equation.
For the quadratic \( u^2 - 23u - 50 \), the discriminant was calculated as:\[ (-23)^2 - 4\times1\times(-50) = 529 + 200 = 729 \]
For the quadratic \( u^2 - 23u - 50 \), the discriminant was calculated as:\[ (-23)^2 - 4\times1\times(-50) = 529 + 200 = 729 \]
- If the discriminant is positive, there are two distinct real roots.
- If the discriminant is zero, there is exactly one real root (a repeated root).
- If the discriminant is negative, there are two complex conjugate roots.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and mathematical operations. In this exercise, we encountered the expression\( x^4 - 23x^2 - 50 \), which can initially seem complex.
The process involved identifying a substitution to rewrite the expression in a simpler quadratic form.Substituting \( u = x^2 \) transformed it into a more familiar expression:
The process involved identifying a substitution to rewrite the expression in a simpler quadratic form.Substituting \( u = x^2 \) transformed it into a more familiar expression:
- Original: \( x^4 - 23x^2 - 50 \)
- Transformed: \( u^2 - 23u - 50 \)
Roots of Equations
The roots of an equation are the solutions to the equation where it equals zero.
For a simple quadratic equation, these roots can be found by factoring, using the quadratic formula, or other algebraic methods. In our exercise, we found that for the quadratic form \( u^2 - 23u - 50 \), the roots were \( u = 25 \) and \( u = -2 \).
By substituting back to \( x^2 = u \), we solved:
For a simple quadratic equation, these roots can be found by factoring, using the quadratic formula, or other algebraic methods. In our exercise, we found that for the quadratic form \( u^2 - 23u - 50 \), the roots were \( u = 25 \) and \( u = -2 \).
By substituting back to \( x^2 = u \), we solved:
- \( x^2 = 25 \) gives roots \( x = 5 \) and \( x = -5 \)
- \( x^2 = -2 \) does not provide real roots, since the square of a real number cannot be negative
