Hi,

If

x² + px + q = (x + a)(x + b)

then q = a * b

and p = a + b

Therefore: if q is positive then a and b have the same sign (both + or both -)

if q is negative then a and b have different signs.

Use these hints to do:

a) x² - 11x + 30 = (x - 5)(x - 6)

b) a² - 7a + 10 = (a - 5)(a - 2)

c) b² + 10b + 24 = (b + 4)(b + 6)

d) y² - 9y + 20 = (y - 5)(y - 4)

e) p² -7p -18 = (p - 9)(p + 2) That's the first problem where the last summand is negative. Therefore the numbers a and b you are looking for must have different signs. The absolute value of the negative number must be greater than the positive one because the coefficient of x is negative.

f) x³ + x² - 6x = x(x² + x - 6) = x(x - 2)(x + 3) Factor out the x. In the bracket is a quadratic term which can be factorized as usual

EB