(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
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
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.