Question

factor completely and solve each equation. 15. $a^{3}-9a^{2}+14a = 0$ 16. $x^{3}=3x^{2}$ 17. $t^{4}-3t^{3}-40t^{2}=0$ 18. $b^{3}-8b^{2}+16b = 0$

Explanation:

Step1: Factor out common factor for equation \(a^{3}-9a^{2}+14a = 0\)

Factor out \(a\): \(a(a^{2}-9a + 14)=0\)

Step2: Factor the quadratic expression

Factor \(a^{2}-9a + 14\) as \((a - 2)(a - 7)\). So the factored - form is \(a(a - 2)(a - 7)=0\)

Step3: Solve for \(a\)

Set each factor equal to zero: \(a=0\) or \(a - 2=0\) or \(a - 7=0\). So \(a = 0\), \(a=2\), \(a = 7\)

Step4: Factor out common factor for equation \(x^{3}=3x^{2}\)

Rearrange to \(x^{3}-3x^{2}=0\), then factor out \(x^{2}\): \(x^{2}(x - 3)=0\)

Step5: Solve for \(x\)

Set each factor equal to zero: \(x^{2}=0\) or \(x - 3=0\). So \(x = 0\), \(x=3\)

Step6: Factor out common factor for equation \(t^{4}-3t^{3}-40t^{2}=0\)

Factor out \(t^{2}\): \(t^{2}(t^{2}-3t - 40)=0\)

Step7: Factor the quadratic expression

Factor \(t^{2}-3t - 40\) as \((t - 8)(t+5)\). So the factored - form is \(t^{2}(t - 8)(t + 5)=0\)

Step8: Solve for \(t\)

Set each factor equal to zero: \(t^{2}=0\) or \(t - 8=0\) or \(t + 5=0\). So \(t = 0\), \(t=8\), \(t=-5\)

Step9: Factor out common factor for equation \(b^{3}-8b^{2}+16b = 0\)

Factor out \(b\): \(b(b^{2}-8b + 16)=0\)

Step10: Factor the quadratic expression

Factor \(b^{2}-8b + 16\) as \((b - 4)^{2}\). So the factored - form is \(b(b - 4)^{2}=0\)

Step11: Solve for \(b\)

Set each factor equal to zero: \(b=0\) or \(b - 4=0\). So \(b = 0\), \(b=4\)

Answer:

For \(a^{3}-9a^{2}+14a = 0\), \(a = 0\), \(a=2\), \(a = 7\); for \(x^{3}=3x^{2}\), \(x = 0\), \(x=3\); for \(t^{4}-3t^{3}-40t^{2}=0\), \(t = 0\), \(t=8\), \(t=-5\); for \(b^{3}-8b^{2}+16b = 0\), \(b = 0\), \(b=4\)