What are the solutions to the equation (4x+1)(x-8)=0 ?

The solutions to the equation (4x+1)(x-8)=0 are x=-1/4 ,x=8

Find the zeros of (4x+1)(x-8)=0

The zeros of (4x+1)(x-8)=0 are x=-1/4 ,x=8

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(4 x + 1) (x - 8) = 0

x = - \frac{1}{4}, x = 8

Decimal Notation

x = - 0.25, x = 8

Solution steps

(4 x + 1) (x - 8) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

4 x + 1 = 0 or x - 8 = 0

Solve

4 x + 1 = 0 : x = - \frac{1}{4}

Solve

x - 8 = 0 : x = 8

The solutions to the quadratic equation are:

x = - \frac{1}{4}, x = 8

\frac{5 !}{2 ! ( 5 - 2 ) !}

\frac{16 a ^{4} - 40 a ^{2} + 24 a}{12 a ^{3}}

328 = 8 (5 - 6 x)

3 \cdot 4

x - 2.41 = 3.42

What are the solutions to the equation (4x+1)(x-8)=0 ?
The solutions to the equation (4x+1)(x-8)=0 are x=-1/4 ,x=8
Find the zeros of (4x+1)(x-8)=0
The zeros of (4x+1)(x-8)=0 are x=-1/4 ,x=8