I'm getting a as 7.75 but am not sure as how to go about further.
in Algebra 1 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

If 3x+5 is a factor then (3x+5)(x+b)(x+c)=(a-1)x³+(a-1)x²-(2a+1)x-15.

That is, (3x+5)(x²+x(b+c)+bc)=(a-1)x³+(a-1)x²-(2a+1)x-15.

3x³+3x²(b+c)+3bcx+5x²+5x(b+c)+5bc=(a-1)x³+(a-1)x²-(2a+1)x-15.

By comparing coefficients:

x³: a-1=3, so a=4; 

x²: 3b+3c+5=a-1=3, 3b+3c=-2, b+c=-2/3;

x: 3bc+5b+5c=-(2a+1)=-9;

constant: 5bc=-15, bc=-3, so -9+5b+5c=-9, 5b+5c=0, b+c=0; but b+c=-2/3, so there’s a contradiction, and therefore no solution. So 3x+5 cannot be a factor of the cubic. The question has been misstated.

 

 

by Top Rated User (717k points)

Related questions

1 answer
1 answer
1 answer
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!
84,491 questions
89,409 answers
1,998 comments
8,085 users