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Question 2

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Factorize into linear factors.

(viii) z4−32z2−3969z^4 - 32z^2 - 3969

Solution

Let u=z2:u2−32u−3969=0Discriminant:Δ=(−32)2−4(1)(−3969)=1024+15876=16900Δ=130Solve for u:u=32±1302u1=32+1302=1622=81u2=32−1302=−982=−49So:z4−32z2−3969=(z2−81)(z2+49)Factor each:z2−81=(z−9)(z+9)z2+49=(z−7i)(z+7i)Therefore:z4−32z2−3969=(z−9)(z+9)(z−7i)(z+7i)\begin{aligned} & \boxed{\text{Let } u = z^2:} \\ \\ & u^2 - 32u - 3969 = 0 \\ \\ & \boxed{\text{Discriminant:}} \\ & \Delta = (-32)^2 - 4(1)(-3969) = 1024 + 15876 = 16900 \\ & \sqrt{\Delta} = 130 \\ \\ & \boxed{\text{Solve for } u:} \\ & u = \frac{32 \pm 130}{2} \\ & u_1 = \frac{32 + 130}{2} = \frac{162}{2} = 81 \\ & u_2 = \frac{32 - 130}{2} = \frac{-98}{2} = -49 \\ \\ & \boxed{\text{So:}} \\ & z^4 - 32z^2 - 3969 = (z^2 - 81)(z^2 + 49) \\ \\ & \boxed{\text{Factor each:}} \\ & z^2 - 81 = (z - 9)(z + 9) \\ & z^2 + 49 = (z - 7i)(z + 7i) \\ \\ & \boxed{\text{Therefore:}} \\ & z^4 - 32z^2 - 3969 = (z - 9)(z + 9)(z - 7i)(z + 7i) \end{aligned}
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