What are the solutions to the equation (k+1)(k-5)=0 ?

The solutions to the equation (k+1)(k-5)=0 are k=-1,k=5

Find the zeros of (k+1)(k-5)=0

The zeros of (k+1)(k-5)=0 are k=-1,k=5

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(k + 1) (k - 5) = 0

k = - 1, k = 5

Solution steps

(k + 1) (k - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

k + 1 = 0 or k - 5 = 0

Solve

k + 1 = 0 : k = - 1

Solve

k - 5 = 0 : k = 5

The solutions to the quadratic equation are:

k = - 1, k = 5

\frac{x ^{- 4} + x ^{2} y ^{- 3}}{( x y ) ^{- 2}}

16 x^{2} - 32 x + 16

\frac{3}{5} t = - 12

(6 v)^{2}

What are the solutions to the equation (k+1)(k-5)=0 ?
The solutions to the equation (k+1)(k-5)=0 are k=-1,k=5
Find the zeros of (k+1)(k-5)=0
The zeros of (k+1)(k-5)=0 are k=-1,k=5