How do you solve #x^2+7x-18=0# by factoring?

Question

Julia Adams · Accepted Answer

While it will not always be the case, start by assuming we are only dealing with integers and the factors of
#x^2+7x-18#
have the form
#(x+a)(x-b)#  we know #a# and #-b# have different signs since their product is negative (#-18#)
We also know that #a# is greater than #b# since the coefficient of #x# is greater than zero. There are only a limited number of possible integer values for #a# and #b# with #a>b# and #ab = 18# #(a,b) = (18,1)# which would give #a-b = 17#; not what we want #(a,b) = (9,2)# which would give #a-b = 7#; this matches the #x# coefficient of the given equation #(a,b)= (6,3)# which would give #a-b=3#; not what we want There are no other integer possibilities. The factoring is
#(x+9)(x-2) = x^2 + 7x -18# Since
#x^2 + 7x - 18 = 0#
either
#x+9 = 0# or #x-2=0# So
either #x=-9# or #x=2#

Hazel Anglin · Answer

undefined To solve the quadratic equation $x^2 + 7x - 18 = 0$ by factoring, follow these steps:

1. Write down the equation: $x^2 + 7x - 18 = 0$.
2. Factor the quadratic expression on the left side of the equation: $x^2 + 7x - 18 = (x - 2)(x + 9)$.
3. Set each factor equal to zero and solve for $x$:
   - $x - 2 = 0$ gives $x = 2$.
   - $x + 9 = 0$ gives $x = -9$.
4. Therefore, the solutions to the equation are $x = 2$ and $x = -9$.