I'm getting a as 7.75 but am not sure as how to go about further.
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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 (711k points)

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