What are the solutions to the equation (u^2-2u-24)(4u-28)=0 ?

The solutions to the equation (u^2-2u-24)(4u-28)=0 are u=-4,u=6,u=7

Find the zeros of (u^2-2u-24)(4u-28)=0

The zeros of (u^2-2u-24)(4u-28)=0 are u=-4,u=6,u=7

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(u^{2} - 2 u - 24) (4 u - 28) = 0

u = - 4, u = 6, u = 7

Solution steps

(u^{2} - 2 u - 24) (4 u - 28) = 0

Factor

(u^{2} - 2 u - 24) (4 u - 28) : 4 (u + 4) (u - 6) (u - 7)

4 (u + 4) (u - 6) (u - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

u + 4 = 0 or u - 6 = 0 or u - 7 = 0

Solve

u + 4 = 0 : u = - 4

Solve

u - 6 = 0 : u = 6

Solve

u - 7 = 0 : u = 7

The solutions are

u = - 4, u = 6, u = 7

x^{3} - a x^{2} - 9 x + 9 a = 0

x^{1} - 3 x^{3} = 8

(x - 3) (x + 1) (x + 2) = 0

- x^{3} - 8 x^{2} + 25 = 0

x^{4} - 10 x^{2} - 3 x + 25 = 0

What are the solutions to the equation (u^2-2u-24)(4u-28)=0 ?
The solutions to the equation (u^2-2u-24)(4u-28)=0 are u=-4,u=6,u=7
Find the zeros of (u^2-2u-24)(4u-28)=0
The zeros of (u^2-2u-24)(4u-28)=0 are u=-4,u=6,u=7