Converting Complex Trinomials into Vertex Form

I need some help factoring part of an equation.

How would I factor -5(t^{2 }- 7t + 49/4). I do believe I have to factor it as a fraction but I have no clue what the fraction should be 2ab should equal -7t and b^{2 }is supposed to equal 49/4.

My guess is.. -5(t - 7/2)^{2} Is that correct?? Since technically 7/2 is 3.5.. And 3.5 x 2 = 7.. And then 7/2^{2 } should equal 49/4.

Re: Converting Complex Trinomials into Vertex Form

Quote:

Originally Posted by

**IvoryChan** I need some help factoring part of an equation.

How would I factor -5(t^{2 }- 7t + 49/4). I do believe I have to factor it as a fraction but I have no clue what the fraction should be 2ab should equal -7t and b^{2 }is supposed to equal 49/4.

My guess is.. -5(t - 7/2)^{2} Is that correct?? Since technically 7/2 is 3.5.. And 3.5 x 2 = 7.. And then 7/2^{2 } should equal 49/4.

Why guess? FOIL it out to check your answer. Which is more or less what you did. So you know you have the correct answer.

-Dan

Re: Converting Complex Trinomials into Vertex Form

I would suggest for easy calculation just take LCM inside the braces and get rid of the denominator.

-5( t^2 - 7t + 49/4 ) = -5 [ ( 4t^2 - 28t + 49 )/4 ] = -5/4 ( 4t^2 - 28t + 49 ) . Now proceed with the steps for splitting the middle term

product of coefficient of x^2 and constant term = 4 * 49 = 2 * 2 * 7 * 7 = 14 * 14 = - 14 * - 14 Now we observe that the sum of these factors is indeed -28

Let us split the middle term

-5/4 ( 4t^2 - 28t + 49 ) = -5/4 ( 4t^2 - 14t - 14 t + 49 ) = -5/4 [ 2t ( 2t - 7 ) - 7 ( 2t - 7 ) ] = - 5/4 ( 2t-7 ) ( 2t - 7 ) = -5/4 ( 2t-7)^2