McDougal Littell Algebra 2, Chapter 5, 5.2, Section 5.2, Problem 19

All you have to do is put in the problem in a calculator and graph:
y = x^2 + 10x + 25

the graph has a zero at -5, therefore this is the p and the q
intercept form is:
y = a (x - p) (x - q)
plug in the number:
y = a ( x - -5) (x - - 5)
y = a (x + 5) (x + 5)
now we have to solve for a. for this we can use any point on the graph. As you can see, the graph crosses the point (-3 , 4) so we can use this coordinate to substitute as x and y
4 = a (-3 + 5) (-3 + 5)
4 = a (2) (2)
4 = 4a
1 = a
so the intercept form is:
y = 1 (x + 5) (x + 5)