+140 Factor \(-16x^4+x^2+2x+1 \) into two quadratic polynomials with integer coefficients. Submit your answer in the form \((ax^2+bx+c)(dx^2+ex+f)\) , with \(a .
Expand the right hand side:
\((-16x^4+x^2+2x+1)=adx^4+aex^3+afx^2+bdx^3+bex^2+bfx+cdx^2+cex+cf\)
\(\hspace{4.4cm}=adx^4+(ae+bd)x^3+(af+be+cd)x^2+(bf+ce)x+cf\)
So:
\(ad=-16 \\ ae+bd=0 \\ af+be+cd=1 \\ bf+ce=2 \\ cf=1 \\\)
The possible combinations of a and d: \((a,d)=(4,-4),(16,-1),(8,-2)\)
\(c=1,f=1\) is a must by the last equation.
So our system becomes:
\(ad=-16 \\ ae+bd=0 \\ a+be+d = 1 \\ b+e = 2 \\\)
It is good idea to try one of the combinations of the above.
Suppose (a,d) = (4,-4)
Then the first equation is satisfied. The second equation is 4e-4b = 0, which only works if e-b=0 or e=b, but from the last equation, b+e = 2 , hence 2e=2 meaning e=1 and thus b=1.
so we solved the system.
Therefore: \(-16x^4+x^2+2x+1=(4x^2+x+1)(-4x^2+x+1)\)
I hope this helps :)!
