(A) I start with writing out the factored form with the coefficients unknown:
[tex]8x^2-14x+5=(ax\pm b)(cx\pm d)[/tex]
We have now to figure out a,b,c, and d. I usually start with the constant term to see in which ways it could be obtained. Clearly, the constant 5 corresponds to the product bd. Well, 5 is prime so this product cannot be anything but 1*5 (or 5*1), so we already have a good choice for b and d:
[tex]8x^2-14x+5=(ax\pm1)(cx\pm5)[/tex]
then I try to see how we could get the -14 in front of x, while keeping an eye on the 8 in front of x^2. Given the b and d values, that can happen as follows: -14 = 5*(-2)+1*(-4), so
[tex]8x^2-14x+5=(2x-1)(4x-5)[/tex]
This is the factored form of the quadratic. So our equation is now
[tex](2x-1)(4x-5)=0[/tex]
This makes solving the equation much easier. When is this product equal zero? When either of the factors equal zero!
[tex](2x-1)(4x-5)=0\iff \\2x-1=0\implies x_1 = \frac{1}{2}\\4x-5=0\implies x_2 = \frac{5}{4}[/tex]
So the solutions are 1/2 and 5/4.
What does this tell you about the graph? The solutions to the quadratic equation are the x-intercepts, so you know that the graph (which is a parabola) intercepts the x axis at 1/2 and also at 5/4.
(B) This is a similar problem but taken in reverse: we start out with known x-intercepts, or solutions to the quadratic, which we are to find. If the solutions are -2 and 1/3, then I can write them in a factored form:
[tex](x+2)(x-\frac{1}{3})=0[/tex]
(please verify for yourself that setting x to either -2 or 1/3 will result in 0)
Now multiply out:
[tex](x+2)(x-\frac{1}{3})=0\\x^2+2x-\frac{1}{3}x-\frac{2}{3}=0\\x^2+\frac{5}{3}x-\frac{2}{3}=0\,\,\,|\cdot 3\\3x^2+5x-2=0[/tex]
which gives you the final quadratic with integer coefficients.
