Question #2d2ff Precalculus Polynomial Functions of Higher Degree Zeros 1 Answer Douglas K. Mar 19, 2017 #q = 6 and q = -6# Explanation: Let #r_1 =# the first root Let #r_2 = # the second root Given: #r_1-r_2=2# #(x - r_1)(x - r_2) = x^2 - qx + 8# #x^2 -r_2x -r_1x + r_1r_2 = x^2-qx+8# #-r_1-r_2 = -q# #r_1-r_2=2# #r_1r_2=8# Substitute for #r_2 = 8/r_1#: #-r_1-8/r_1 = -q# #r_1-8/r_1=2# #q = r_1+8/r_1" [1]"# #r_1^2-2r_1-8=0" [2]"# Factor equation [2]: #(r_1-4)(r_1+2)# #r_1 = 4 and r_1 = -2# #q = 4 + 8/4 and q = -2+8/-2# #q = 6 and q = -6# Answer link Related questions What is a zero of a function? How do I find the real zeros of a function? How do I find the real zeros of a function on a calculator? What do the zeros of a function represent? What are the zeros of #f(x) = 5x^7 − x + 216#? What are the zeros of #f(x)= −4x^5 + 3#? How many times does #f(x)= 6x^11 - 3x^5 + 2# intersect the x-axis? What are the real zeros of #f(x) = 3x^6 + 1#? How do you find the roots for #4x^4-26x^3+50x^2-52x+84=0#? What are the intercepts for the graphs of the equation #y=(x^2-49)/(7x^4)#? See all questions in Zeros Impact of this question 1089 views around the world You can reuse this answer Creative Commons License