



 
Hi Mukulu, If all the polynomials have common factor, we can find the common factor(s) of the first two and use the remainder theorem to find the value of "a". If I look at 2x^{2}x6 it can broken into factors of 2x^{2}x6 = 2x^{2}4x+3x6 = 2x(x2)+3(x2)=(2x+3)(x2) Since 2x^{2}x6 has factors of (2x+3)& (x2) we know it has roots at x=3/2 & x=2 and the other two polynomials must at least one of these roots. So let's use the remainder theorem to find if x=3/2 or x=2 are roots of 3x^{2}8x+4 f(3/2)=3(3/2)^{2}8(3/2)+4= 22.75 f(2)=3(2)^{2}8(2)+4= 0 Since x=2 yields a remainder of 0 for 3x^{2}8x+4 we know that (x2) is the only common factor for all of the polynomials. If you substitute x=2 into your last equation and set it equal to 0, you can solve for a using simple algebra. Hope this helps, Janice  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 