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Ch. P - Fundamental Concepts of Algebra
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 1, Problema 79

In Exercises 65–92, factor completely, or state that the polynomial is prime. x3+2x2−4x−8

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1
Group the terms of the polynomial into two pairs: \( (x^3 + 2x^2) \) and \( (-4x - 8) \). This is called factoring by grouping.
Factor out the greatest common factor (GCF) from each group. From \( (x^3 + 2x^2) \), the GCF is \( x^2 \), so it becomes \( x^2(x + 2) \). From \( (-4x - 8) \), the GCF is \( -4 \), so it becomes \( -4(x + 2) \).
Notice that both groups now contain the common factor \( (x + 2) \). Factor \( (x + 2) \) out of the entire expression: \( x^2(x + 2) - 4(x + 2) = (x + 2)(x^2 - 4) \).
Recognize that \( x^2 - 4 \) is a difference of squares. Use the formula \( a^2 - b^2 = (a - b)(a + b) \) to factor \( x^2 - 4 \) into \( (x - 2)(x + 2) \).
Combine all the factors to write the completely factored form of the polynomial: \( (x + 2)(x - 2)(x + 2) \). Simplify if necessary.

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