y=-x^2+4x+3

Question

this question have to change into factored form

Answer 1

To answer this question you should use the quadratic formula.

-b±√(b^2-4ac)

         2

ax^2+bx+c

So a=-1 b=4 and c=3

Fill in the quadratic formula:

-4±√((-4)^2-4(-1)(3))

           2(-1)

Simplify:

-4±√(16+12)

       -2

-4±√(28)

    -2

You can pull a 4 out of 28 to make it:

-4±2√(7)

   -2

Simplify again:

2±√(7)*

*Note: you can divide by -2 and the plus or minus sign still "applies."

That gives us:

-x^2+4x+3 = (-x+(2+√(7))(x-(2-√(7))

Now, double check by FOILing the factored form:

(-x+(2+√(7))(x-(2-√(7)) = -x^2+(2-√(7))x+(2+√(7))x-(2+√(7))(2-√(7))

=-x^2+2x-√(7)x+2x+√(7)x-(4-2√(7)+2√(7)-7)

=-x^2+2x+2x+√(7)x-√(7)x-(4-7)
=-x^2+4x-(-3)

=-x^2+4x+3

So the answer is definitely: (-x+(2+√(7))(x-(2-√(7))

Answer 2

I think a technique called "completing the square" is needed here.  It rewrites

y = Ax² + Bx + C in the form y = A(x + P)² + Q.

It is accomplished in several steps.  To demonstrate, I'll use an example like your own, leaving it for you to try yourself once you learn the method.

Say we want to "complete the square" on 

y = 2x² + 8x - 5.

Step 1: Factor out the A=2 from just the x2 and x terms.

y = 2x² + 8x - 5 = 2( x² + 4x ) - 5.

y = 2( x² + 4x ) - 5.

Step 2: Now calculate half of the x term coefficient and square it.

y = 2( x² + 4x ) - 5.

( + 4 / 2 )² = ( 2 )² = 4.

Step 3: Add and subtract the value calculated above inside the ()'s.

y = 2( x² + 4x + 4 - 4 ) - 5.

Step 4: Now a perfect square lies inside the ()'s.

y = 2( x² + 4x + 4 - 4 ) - 5 = 2[ ( x + 2 )² - 4 ] - 5.

[check this for yourself by expanding ( x + 2 )².]

Step 5: Finish by distributing the A and simplifying.

y = 2[ ( x + 2 )² - 4 ] - 5.

y = 2( x + 2 )² - 2(4) - 5 = 2( x + 2 )² - 8 - 5.

y = 2( x + 2 )² - 13.

"completing the square" has rewritten

y = 2x² + 8x - 5 in the form y = 2( x + 2 )² - 13.

Try following this process on your example, and see how it works for you.

I hope this helps.  If you need more help, I or another tutor will be happy to assist you.