can you show how to factor this problem please show the work?

This is not a factorising problem.

This is a quadratic equation, to be solved by the standard formula

The form of quadratic equation is:

az^2 - bz + c = 0 where a = 2, b = -5, c = 2

The formula for solving quadratic equation is z = (-b+SQRT(b^2-4ac))/2a And z = (-b-SQRT(b^2-4ac))/2a

Substituting the values of a, b and c into the formula

z = (5 + SQRT((-5)^2-4*2*2))/2*2 And 5 - SQRT((-5)^2-4*2*2))/2*2

z = (5 + SQRT((25-16))/4 And (5 - SQRT((25-16))/4

z = (5+3)/4 And (5-3)/4

Therefore there are 2 solutions to the given equation z = 2 and z = 0.5

To validate: substitute 0.5 into equation, 2*0.25 + 2 - 5*0.5 = 0

And 2*2^2 + 2 - 5*2 = 0

Validated

2z^2 = -2 + 5z

2z^2 - 5z + 2 = 0

(2z - 1)(z - 2) = 0

2z - 1 = 0

2z = 1

z = 0.5

z - 2 = 0

z = 2

You can see that there is a 2z^2 in there, the only way this factorises nicely is if you put 2z in one of the brackets, and z in the other, so you have:

(2z )(z )

Next you can see that the middle term -5z is negative, while the last term +2 is positive. The only way that can happen is if there are two minus' in the brackets, so now you have:

(2z - )(z - )

Now you have to find factors for the last term, since it's a 2, the only factors are 1 and 2, so you have to stick them in the brackets in such a way that you can get -5z when you multiply them out. In this case it's -2*2z - 1*z = -5z, so you drop the, in like so:

(2z - 1)(z - 2)

And you're done.

Put it into the standard form

2z^2 - 5z + 2 = 0

factors will be

(pz +q)(rz +s)

p x r = 2 (so will be 1 and 2 or 2 and 1) or negatives of these

q x s = 2 (so will also be 1 and 2 or 2 and 1) or negatives of these

the - 5 in the middle is made up of (p x s) + (q x r)

Given the options of values for p, q, r, s it looks as if a (2 x -2) and a (1 x -1) will give us -5

so (2z - 1)(z - 2) looks like a good answer.

Add 2 and -5z to both sides of the equation

2z^2 - 5z + 2 = 0

(2z - 1)(z - 2) = 0

If you wish to solve:

z = 1/2, 2

2z^2 = -2 + 5z

rewrite as:

2z^2 - 5z + 2 = 0

(2z -1) (z - 2) = 0

z = 1/2 or z = 2

My own popular technique for such factoring is the increasing and grouping technique. you're taking the 1st coefficient and multiply it contained in the final coefficient (whilst contained in the order given above). 3*-8=-24 Now you prefer to locate 2 elements of -24 that upload as much as -2 -6 and -4 artwork, so we use those as a substitute of -2 you pass from 3x^2-2x-8 to 3x^2-6x+4x-8 Now, contained in the 1st area of the equation, you have 3x^2-6x, which could be written as 3x(x-2) And the 2d area, 4x-8, could be written as 4(x-2). observe the (x-2) ingredient is the comparable. the certainty that those are the comparable serves in itself as a examine to permit you be conscious of that it is factorable and your math so a techniques is actual. Now you have the 1st area 3x(x-2) and 2d area, 4(x-2), and additionally you're including them, so which you have 3x(x-2)+4(x-2) you have a difficulty-loose ingredient of (x-2), so which you employ distributive components and get (3x+4)(x-2) it incredibly is your very final answer. in case you have been to clean up the equation, and have been instructed it incredibly is comparable to 0, you will possibly set each and every ingredient equivalent to 0 and get x=-4/3 and x=2 as strategies. simply by fact the subject is calling in simple terms for factoring, in spite of the indisputable fact that, you pass away it as (3x+4)(x-2). endure in strategies this basically works for trinomials wherein the 2d time era (x) has 0.5 the flexibility of the 1st time era (x^2)