Verified answer

Ok...

So this is the difference of two squares, what you do first is multiply the number in-front of your x^2 by the last number in the equation. 2 x 1 = 2

So now think of the multiples of two, it only is 1 and 2, if you +_ these will they give minus three? Yes, -2 - 1= -3 .

So, we know then that the -3x can be split into -2x and -x giving your equation now after splitting the middle term :

y = 2x^2 -2x -x +1

So in the first to x's take a 2x out of both:

y = 2x (x - 1) -x +1

Now think how do I get the last two to be the same as the bracket (x - 1)? What can I take out? -1 is the answer so:

y = 2x (x - 1) -1 (x - 1)

Now we have two of the same bracket, we then know one bracket of the answer is (x - 1) the other is made up of the other two terms outside the bracket, 2x and -1. There for the fully factorized answer is:

y = (2x - 1)(x - 1)

Hope this helped

x

Factors of 2 are: 2 and 1. So the brackets will look like this:

(2x + ???)(x + ???)

We need two numbers such that 2*a + 1*b = -3 and a*b = 1

Since a*b = 1, the only possible option is 1*1.

To obtain a -3 in the middle, we must make both the 1's negative, since -1 * -1 = 1

So it should look like this:

(2x - 1)(x - 1)

You can test it by expanding if you are not sure.

y = (2x - 1)(x - 1)

Just a note to be careful about signs and reasoning with respect to Chuck Norris' solution:

"Since a*b = 1, the only possible option is 1*1."

a*b = 1 implies a and b are both 1 OR a and b are both -1

For a and b both 1, the x term would be 3x, but as CN points out a and b must both be -1.

My extreme college calc instructor taught me a candy rhyme to undergo in recommendations the quotient and product rule: Product Rule: 2d "d" first + first "d" 2d meaning: 2d ingredient cases the derivative of the 1st, plus the 1st ingredient cases the derivative of the 2d. Quotient Rule: appropriate "d" backside minue backside "d" appropriate over the backside one squared (Dont comprehend in case you like that yet i admire those, they help me undergo in recommendations soo lots) in any case ok so for the 1st one, you have (2x+a million)(3x-2)^4 so 2d "d" first: (3x-2)^4 * 2 ----> the derviative of 2x +a million is in simple terms 2 plus first "d" 2d: (2x+a million)* 4(3x-2)^3*3 the derivative of (3x-2)^4 is you handle the interior the brackets as variable, so which you convey down the exponenent, and then subtract one from it, 4(3x-2)^3 after for you to multiply it by way of the derivative of whats interior the brackets so 3 => 4(3x-2)^3*3 = (3x-2)^4 * 2 + (2x+a million)* 4(3x-2)^3*3 =2 [ (3x-2)^4 +6(2x + a million)*(3x-2)^3 ] ---> take out a ingredient of two =2(3x-2)^3 [ (3x-2)+ 12x + 6] ---> take out a uncomplicated ingredient of (3x-2)^3 =2(3x-2)^3 [ 15x +4]

y = ( 2x - 1 ) ( x - 1 )

y = (2x-1)(x-1)