Question

PLEASE HELP! Is the following trinomial factorable why or why not h=-4(4t^(2)-9t+2)

Answer 1

Answer:

• h = -4(t - 2)(4t - 1)

Step-by-step explanation:

Given trinomial

• h= -4(4t² - 9t + 2)

Let's try to factor 4t² - 9t + 2

First find it's zeros:

• t = (9 ± √9² - 4*4*2)/(2*4)
• t = (9 ± √81 - 32)/8
• t = (9 ± 7)/8
• t = 2 and t = 1/4

So yes, it is factorable:

• 4t² - 9t + 2 = 4(t - 2)(t - 1/4)
• or
• 4t² - 9t + 2 = (t - 2)(4t - 1)

The given trinomial factored:

• h = -4(t - 2)(4t - 1)

Answer 2

Answer:  Yes, this trinomial -4(4t² -9t + 2) is factorable.

It is   h =(-4)(4t -1 )(t -2) converted from the explanation

$y\ =\left(-4\right)\left(4x\ -1\right)\left(x\ -2\right)$

A graph of the equation is in the attachment. There are two equations, but only one graphed parabola because they are identical.

Step-by-step explanation: I am using x in place of t in this explanation, and leaving out the factored (-4) to simplify.

Using the systematic new AC Method to factor trinomials (Socratic Search)  Starting with a quadratic equation in the form

y = ax² + bx + c  

This involves converting  the equation by multiplying the leading coefficient, a, times the constant, c

y =4x² –9x + 2

=4(x + p)(x + q)

Converted trinomial:

y' = x² –9x + 8

= (x + p')(x + q').

Here p' and q' will have the same – signs because the b term is negative and the c term is positive.

Factor pairs of ac = 8 are -1 and –8. This sum is –9 = b. Then p' = -1 and q' = -8.

Then divide the second factor by the original coefficient, 4.

Replacing the original leading factor (-4) the factored equation looks like

y = -4(4x - 1)(x - 2)

To solve for x

p =p’/a  = ¼   or 4x - 1 = 0  4x = 1   x = ¹/₄

and

q ='q/a

= 8/4 = 2    OR  x -2 = 0  x = 2

The second attachment shows some other ways to set up the equation without the (-4) to solve, and then replace that factor. You get the same values for x.

I hope this helps and is not too confusing. There may be other ways to factor this equation.