# (x+1)(2x-5)=0Not sure if I am heading in the right direction by x(2x-5)+1(2x-5)=0 2x^2-5x+1x-5=0

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I think you have made a few mistakes in your question and I have corrected them.

Now you have given that (x+1)(2x - 5) =0

This is sufficient to find the values of x.

You can merely equate x+1 to 0 and 2x - 5 to 0.

x+1 = 0

=> x = -1

and 2x - 5 =0

=> 2x = 5

=> x = 5/2

You don't need to open the brackets and multiply to arrive at the original equation again.

Of course, if your aim is to multiply (x+1) and (2x -5) and find a quadratic equation it would be

(x+1)(2x - 5) =0

=> 2x^2 - 5x + 2x - 5 =0

=> 2x^2 - 3x - 5 = 0.

We'll start from the fact that a product is zero if one of the factors is zero.

According to this, the product

(x+1)(2x-5) = 0 if and only if:

x+1 = 0

or

2x - 5 = 0

We'll solve the first equation:

x + 1 = 0

We'll subtract 1:

x = -1

We'll solve the second equation:

2x-5 = 0

We'll add 5 both sides;

2x = 5

We'll divide by 2:

x = 5/2

**So, the product is zero if and only if x has the values:**

**x = -1 or x = 5/2.**

Hope you want to solve the equation (x+1)(2x-5) = 0.

We solve the equation to know the value of the variable for which the equation is satisfied.

If a product a*b = 0, the a = 0 or b = 0. this is the law of zer product.

Since (x+1)(2x - 5) = 0, then by the law of zero product x+1 = 0 or 2x-5 = 0.

x+1 = 0 gives x= -1.

2x-5 = 0 gives 2x= 5, or x= 5/2 = 2.5.

Therefore x = -1 or x = 2.5 are solution to (x+1)(2x-5) = 0.

What did is writing the equation (x+1)(2x-5) = 0 in the expanded form . But here is an error.

(x+1)(2x-5) = x(2x-5)+1(2x-5).

(x+1)(2-5) = 2x^2-5x+2x-5.

(x+1)(2x-5) = 2x^2 - 3x-5.

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