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 (81.5k points)

Related questions

1 answer
asked Nov 18, 2012 in Algebra 2 Answers by anonymous | 426 views
1 answer
asked Aug 20, 2012 in Calculus Answers by anonymous | 261 views
1 answer
asked May 5, 2014 in Algebra 1 Answers by kidio Level 1 User (140 points) | 367 views
1 answer
asked Nov 26, 2012 in Calculus Answers by anonymous | 249 views
1 answer
asked Oct 13, 2012 in Algebra 2 Answers by anonymous | 393 views
1 answer
asked Aug 20, 2012 in Algebra 2 Answers by anonymous | 283 views
1 answer
1 answer
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!
85,260 questions
90,519 answers
2,143 comments
81,870 users