What are the solutions to the equation (x^2+6x-7)(2x^2-5x-3)=0 ?

The solutions to the equation (x^2+6x-7)(2x^2-5x-3)=0 are x=1,x=-7,x=3,x=-1/2

Find the zeros of (x^2+6x-7)(2x^2-5x-3)=0

The zeros of (x^2+6x-7)(2x^2-5x-3)=0 are x=1,x=-7,x=3,x=-1/2

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(x^{2} + 6 x - 7) (2 x^{2} - 5 x - 3) = 0

x = 1, x = - 7, x = 3, x = - \frac{1}{2}

Decimal Notation

x = 1, x = - 7, x = 3, x = - 0.5

Solution steps

(x^{2} + 6 x - 7) (2 x^{2} - 5 x - 3) = 0

Factor

(x^{2} + 6 x - 7) (2 x^{2} - 5 x - 3) : (x - 1) (x + 7) (x - 3) (2 x + 1)

(x - 1) (x + 7) (x - 3) (2 x + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x + 7 = 0 or x - 3 = 0 or 2 x + 1 = 0

Solve

x - 1 = 0 : x = 1

Solve

x + 7 = 0 : x = - 7

Solve

x - 3 = 0 : x = 3

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

The solutions are

x = 1, x = - 7, x = 3, x = - \frac{1}{2}

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

3 x^{3} + 1 5^{2} - 3000 = 0

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

2 a^{3} + 0.5 (4 a + 8) = 9 - 2 a

0 = 2 x^{3}

What are the solutions to the equation (x^2+6x-7)(2x^2-5x-3)=0 ?
The solutions to the equation (x^2+6x-7)(2x^2-5x-3)=0 are x=1,x=-7,x=3,x=-1/2
Find the zeros of (x^2+6x-7)(2x^2-5x-3)=0
The zeros of (x^2+6x-7)(2x^2-5x-3)=0 are x=1,x=-7,x=3,x=-1/2