Mathematics, 15.01.2021 01:10 stastnylindsey

To factor a $x^{2}$ +bx+c, you have to group it as (a $x^{2}$ +ux)+(vx+c) so that uv=ac and u+v=b. Both addition and multiplication are commutative, so how do you determine which value is assigned to u and which to v? Or are they interchangeable?

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Answer from: dakshshberry

I'm not sure. Try asking Google.

Pl give me Brainliest

Answer from: itsjoke5550

They are interchangeable

For example, let's expand out (x+3)*(2x+1). Use of the FOIL rule should get to you 2x^2 + 7x + 3.

Now imaging that you handed 2x^2 + 7x + 3 to a friend to be factored. They do not know the factorization yet.

They would see that a = 2, b = 7, c = 3.

The friend needs to find two numbers u,v such that they multiply to a*c = 2*3 = 6 and add to b = 7

Through trial and error (I recommend making a table of values), your friend would get

u*v = 1*6 = 6u+v = 1+6 = 7

But as you mentioned, your friend could easily swap the order of addition and multiplication to say

u*v = 6*1 = 6u+v = 6+1 = 7

If we go with the first option, then the factor by grouping looks like this

2x^2 + 7x + 3

2x^2 + 1x + 6x + 3 ... break up 7x into 1x+6x

(2x^2+1x) + (6x+3) group up the terms into pairs

x(2x+1) + 3(2x+1) factor each grouping

(x+3)(2x+1) factor out the overall GCF

Or we could have the factor by grouping looks like this

2x^2 + 7x + 3

2x^2 + 6x + 1x + 3 ... break up 7x as 6x+1x

(2x^2+6x) + (1x+3)

2x(x+3) + 1(x+3)

(2x+1)(x+3)

(x+3)(2x+1)

We end up with the same factorization.

Answer from: Quest

translate k to n and reflect across the line containing jk.

Answer from: Quest

center (1,3) with radius 6

step-by-step explanation:

we must complete the square to find the center and radius of the circle.

first make sure the x and y squared terms have 1 as their coefficients. we also make sure x and y terms together.

$x^2-2x+y^2-6y=26$

we now create space between the x and y terms with parenthesis.

$(x^2-2x)+(y^2-6y)=26$

we complete the square by taking the middle terms -2x and the -6y - divide each and square them.

$\frac{-2}{2} =(-1)^{2} =1$

$\frac{-6}{2} =(-3)^{2} =9$

we add the squares to both sides.

$(x^2-2x+1)+(y^2-6y+9)=26+1+9$

simplify.

$(x^2-2x+1)+(y^2-6y+9)=36$

and write the quadratics in factored form.

$(x-1)^{2} +(y-3)^{2} =36$

the center is (h,k) or (1,3). the radius is the square root of 36 which is 6.

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To factor a+bx+c, you have to group it as (a+ux)+(vx+c) so that uv=ac and u+v=b. Both addition and m...