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@sshayer
\[g \prime(x)=-2(x-7)\] g(x) is increasing if g'(x)>0, -2(x-7)>0 x-7<0 x<7 including end points if x<=7 g(x) is decreasing if g'(x)<0 -2(x-7)<0 x-7>0 x>7 including end points g(x) is decreasing if x>=7
you know derivatives.
you followed how i solved.
imaginary roots always occur in pairs. Hence zeros are 5,-3,-1+3i,-1-3 i reqd. polynomial f(x)=(x-5)(x+3){x-(-1+ 3 i)}{x-(-1-3 i)} f(x)=(x-5)(x+3){(x+1)-3 i}{(x+1)+3 i} =(x-5)(x+3){(x+1)^2-(3 i)^2} =(x^2-5x+3x-15){x^2+2x+1-9 i^2} =(x^2-2x-15)(x^2+2x+1-9*-1} =(x^2-2x-15)(x^2+2x+1+9) =(x^2-2x-15)(x^2+2x+10) =? simplify further.
Please review what you've done and what you've learned. What do you understand and what do you still need help with? Have you found the "critical values" of this function? To do that, find the first derivative of the function, set the result = to zero, and solve for x. Please show your work.
\[(x^2-2x-15)(x^2+2x+10)=x^4+2x^3+10x^2-2x^3-4x^2-20x-15x^2-30x-150\] -20x-150 =\[=x^4-9x^2-50x-150\]
\[=x^4+2x^3+10x^2-2x^3-4x^2-20x-15x^2-30x-150\] \[=x^4-9x^2-50x-150\]
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