Factor this: p(x) = x^3-6x^2+12x-8 and refer the solution to the attached image

Algebra and Trigonometry (6th Edition)
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ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Factor this: p(x) = x^3-6x^2+12x-8 and refer the solution to the attached image
5. p(x) = x-4x2+ x+ 6,
* This method of factoring involves factoring the coefficients of the leading
term and the constant term into its prime factors.
* The first term x'whose coefficient is 1 can be factored into (x) (x) (x) & the constant term 6 into (1) (2)(3).
Then form three possible binomial factors, (x+1 )(x- 2 )(x-3). Then check each binomial factor by
p(-1) = (- 1) – 4(-1}? + (-1) +6
= - 1-4-1+6
applying the Factor Theorem. Test for ( x + 1), x-c=x+ 1, c=- 1,
Since p (-1 ) = 0, then (x-1) is a factor.
= -6 +6 =0
p( 2) = (2)-4(2)+2 +6
= 8- 16 +2 +6
Test for ( x-2), x-c =x-2, c 2,
= 16-16 0
Since p (2)=0, then (x-2)is a factor.
Transcribed Image Text:5. p(x) = x-4x2+ x+ 6, * This method of factoring involves factoring the coefficients of the leading term and the constant term into its prime factors. * The first term x'whose coefficient is 1 can be factored into (x) (x) (x) & the constant term 6 into (1) (2)(3). Then form three possible binomial factors, (x+1 )(x- 2 )(x-3). Then check each binomial factor by p(-1) = (- 1) – 4(-1}? + (-1) +6 = - 1-4-1+6 applying the Factor Theorem. Test for ( x + 1), x-c=x+ 1, c=- 1, Since p (-1 ) = 0, then (x-1) is a factor. = -6 +6 =0 p( 2) = (2)-4(2)+2 +6 = 8- 16 +2 +6 Test for ( x-2), x-c =x-2, c 2, = 16-16 0 Since p (2)=0, then (x-2)is a factor.
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