Factoring Question?

Question

How would you express this quadratic expression x^2-(2√3+1)x+2√3 in factored form using the quadratic formula?

The answer is (x-2√3)(x-1). Could anyone show me the process work?

Answer 1

Even without the quadratic formula you should be able to see the following:

x² - (2√3 + 1)x + 2√3

Distribute the x through the parentheses:
x² - 2√3x - x + 2√3

Now group the terms:
x² - x - 2√3x + 2√3

From the first two terms you can distribute out an x:
x(x - 1) - 2√3x + 2√3

And from the second two terms you can distribut out -2√3
x(x - 1) - 2√3(x - 1)

Finally you can distribute out the common (x - 1):
(x - 2√3)(x - 1)

I'll explain the quadratic formula method if you like...

Answer 2

The middle term (2sqrt3 +1) is a big clue toward the factoring.
Does 2sqrt3 * 1 = the final term? Yes, then the factors are
2sqrt 3 and 1

Since the middle term is negative and the final term is positive, you have x - something times x - something.

(x - 2sqrt3)(x-1) works.

Answer 3

quadratic formula says:
x = (-b±√[b²-4ac]) / (2a)
So using your numbers gives:
(2√3+1±√[(2√3+1)²-4*1*2√3]) / 2
=(2√3+1±√[(12+4√3+1)-8√3]) / 2
=(2√3+1±√[(12-4√3+1)]) / 2
=(2√3+1±√[(2√3-1)²]) / 2
=(2√3+1± (2√3-1)) / 2
=(2√3+1+(2√3-1)) / 2 or =(2√3+1 - (2√3-1)) / 2
=(4√3)/2 or 2/2
= 2√3 or 1
So in factorise form it is:
(x-2√3)(x-1)

Answer 4

This is the formula:
ax^2 + bx + c = 0 and x1 and x2 are the solutions
--> ax^2 + bx + c = a(x-x1)(x-x2)

x^2 - (2sqrt3 + 1)x + 2sqrt3 = 0
delta = (2sqrt3+1)^2 - 8sqrt3 = (2sqrt3-1)^2

x1 = (1/2)[2sqrt3+1-2sqrt3+1]=1
x2 = (1/2)[2sqrt3+1+2sqrt3-1]=2sqrt3

--> (x-1)(x-2sqrt3)

Answer 5

when you have an expression written as ax^2 + bx + c (quadratic form), you can factor it as such:

(a1x + b1)(a2x + b2)

as long as a1*a2 = a, a1*b2 + a2*b1 = b, and b1*b2 = c

So our a = 1, our b = -(2 sqrt(3) + 1), our c = 2 sqrt(3)

so...
a1 and a2 are 1 (1*1 = 1)

So we need a b1 and b2 such that b1 + b2 = -2 sqrt(3) - 1

therefore b1 = -2 sqrt(3) and b2 = -1

(1x + -2 sqrt(3)) (1x + -1)
(x - 2 sqrt(3))(x - 1)

Hope this makes sense! :) Best of luck.