Factor 2x^2-x-10. Rewrite the trinomial with the x term expanded using 2 factors.Then group the first 2 and last 2 terms together and find the Gcf of each Trinomial (2x- ___) + (4x-10)
Gcf __ (2x-5) + __ (2x-5)

Trinomial: (2x² – 5x) + (4x – 10)

GCF: x (2x – 5) + 2 (2x – 5)

Step-by-step explanation:

2x² – x – 10

The equation above can be factorised as follow:

2x² – x – 10

Multiply the first term (i.e 2x²) and the last term (i.e –10) together.

2x² × (–10) = –20x²

Find two factors of –20x² such that when we add them together, it will result to the 2nd term in the equation (i.e –x).

The factors are 4x and –5x.

Next, replace –x in the equation above with 4x and –5x. This is illustrated below:

2x² – x – 10

2x² – 5x + 4x – 10

Next, factorise

2x² – 5x + 4x – 10

Trinomial: (2x² – 5x) + (4x – 10)

GCF: x (2x – 5) + 2 (2x – 5)

see below

Step-by-step explanation:

A.  f(x) = 4x^2 - 7x - 15

We need the numbers to multiply to- 15

1*15  and 3*5  

One of the terms multiplies by 4 (or 2)

Try 4 first

4*-3 +5 = -7   That works

-3+5 = -15

f(x) = (4x+5)(x-3)

B.  We want the x intercepts so set equal to 0

0 = (4x+5)(x-3)

Using  the zero product  property

0 = 4x+5    x-3 =0

-5 = 4x           x=3

-5/4 =x            x=3

C  End behavior

Let x be negative infinity

The function is dominate by 4x^2

f(-∞) = 4 ( -∞) ^2 = 4 (∞) = ∞  

At negative infinity the functions goes to infinity

Let x be  infinity

The function is dominate by 4x^2

f(∞) = 4 ( ∞) ^2 = 4 (∞) = ∞  

At  infinity the functions goes to infinity

Part D:

We know the zeros at 3 and -5/4

We know the vertex is at  halfway between the zeros

(3 -5/4) /2 =  7/4 /2 = 7/8

Since the parabola opens up (The coefficient of the x^2 term is positive), we know we have a minimum.

f(7/8) = 4 (7/8)^2 -7(7/8) -15 = 4(49/64) -1/8 -15 =-289/16

We know the minimum the zeros and the end behavior, we can graph the functions