Thus, $ f(x^2 - 3) = 2(x^2 - 3)^2 - 2(x^2 - 3) - 1 = 2(x^4 - 6x^2 + 9) - 2x^2 + 6 - 1 = 2x^4 - 12x^2 + 18 - 2x^2 + 5 = 2x^4 - 14x^2 + 23 $. - IQnection
Title: Simplify $ f(x^2 - 3) = 2(x^2 - 3)^2 - 2(x^2 - 3) - 1 $: Full Expansion & Insights
Title: Simplify $ f(x^2 - 3) = 2(x^2 - 3)^2 - 2(x^2 - 3) - 1 $: Full Expansion & Insights
Meta Description:
Learn how to simplify $ f(x^2 - 3) = 2(x^2 - 3)^2 - 2(x^2 - 3) - 1 $. This step-by-step breakdown reveals the fully expanded form $ 2x^4 - 14x^2 + 23 $, helping you master function substitution and algebraic simplification.
Understanding the Context
Introduction
Working with function compositions and algebraic expressions can be challenging, especially when expanding nested expressions like $ f(x^2 - 3) $. In calculus, algebra, or applied math, expressing a function in terms of a transformed variable enables deeper understanding of behavior, graphs, and optimization.
In this article, we simplify
$$
f(x^2 - 3) = 2(x^2 - 3)^2 - 2(x^2 - 3) - 1
$$
by expanding it fully into a polynomial in terms of $ x $, resulting in
$$
2x^4 - 14x^2 + 23.
$$
Understanding this process not only aids in solving equations but also strengthens your skills in polynomial manipulation, substitution, and function theory.
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Key Insights
Step-by-Step Expansion
Start with the original expression:
$$
f(x^2 - 3) = 2(x^2 - 3)^2 - 2(x^2 - 3) - 1
$$
Step 1: Expand $ (x^2 - 3)^2 $
$$
(x^2 - 3)^2 = x^4 - 6x^2 + 9
$$
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Multiply by 2:
$$
2(x^2 - 3)^2 = 2x^4 - 12x^2 + 18
$$
Step 2: Expand $ -2(x^2 - 3) $
$$
-2(x^2 - 3) = -2x^2 + 6
$$
Step 3: Combine all terms
Now combine every part:
$$
f(x^2 - 3) = (2x^4 - 12x^2 + 18) + (-2x^2 + 6) - 1
$$
Add constants and like terms:
- $ 2x^4 $ stays
- $ -12x^2 - 2x^2 = -14x^2 $
- $ 18 + 6 - 1 = 23 $
Result:
$$
f(x^2 - 3) = 2x^4 - 14x^2 + 23
$$
Why This Matters
Simplifying expressions like $ f(x^2 - 3) $ into standard polynomial form reveals key features:
- Degree of the polynomial: The highest exponent is 4, so $ f(x^2 - 3) $ is degree-4 in $ x $.
- Symmetry: Since only even powers appear, the function is even in terms of $ x^2 $.
- Roots and behavior: Expanded form helps locate zeros, plot graphs, and analyze curvature.
