Browse · MATH Print → jmc algebra junior Problem Solve for the largest value of x such that x2+24x+128=0. Solution — click to reveal Factoring, we have x2+24x+128=(x+16)(x+8)=0. Hence, the possible values for x are −16 and −8, and the larger value of the two is −8. Final answer -8 ← Previous problem Next problem →