Factoring of equation in Pre cal
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

factor p(x)=1/8[(2x^4)+(3x^3)-16x-24)]^2

By inspection, p(x) = 0 for x = 2.

Thus p(x) has a factor of (x - 2). This gives,

p(x) = 1/8[{x - 2}{2x^3 + 7x^2 + 14x +12}]^2

Further inspection shows p(x) decreasing in value as x -> -2, from below, and increasing in value as x becomes greater than -1, suggesting a minimum value in the ramge [-2 < x < -1].

Closer inspection shows a zero at x = -1.5, ie there is a factor of (2x + 3).

p(x) = 1/8[{x - 2}{2x + 3}{x^2 + 2x + 4}}]^2

The final quadratic is unfactorisable. Hence the final answer is:

p(x) = 1/8[{x - 2}{2x + 3}{x^2 + 2x + 4}}]^2‚Äč

by Level 11 User (80.8k points)

Related questions

1 answer
asked Nov 18, 2012 in Algebra 2 Answers by anonymous | 382 views
1 answer
asked May 5, 2014 in Algebra 1 Answers by kidio Level 1 User (140 points) | 289 views
1 answer
asked Aug 20, 2012 in Calculus Answers by anonymous | 227 views
1 answer
asked Nov 26, 2012 in Calculus Answers by anonymous | 211 views
1 answer
1 answer
asked Oct 13, 2012 in Algebra 2 Answers by anonymous | 331 views
1 answer
asked Aug 20, 2012 in Algebra 2 Answers by anonymous | 236 views
1 answer
1 answer
asked Sep 5, 2014 in Other Math Topics by nrmorgan Level 1 User (120 points) | 475 views
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
84,043 questions
88,973 answers
1,990 comments
6,686 users