How do you factor q^2 - 7q - 10?

2 Answers
Sep 28, 2015

x_1= 7/2+1/2*(sqrt(89))
x_2= 7/2-1/2*(sqrt(89))

Explanation:

quadratic formula
x_1,_2 =(-b+- sqrt(b^2-4ac))/(2a)
here a=1,b=-7,c=-10
so
x_1 =(-(-7)+ sqrt((-7)^2-4*1*(-10))/(2*1))
= 7/2+1/2*sqrt(89)
x_2 =(-(-7)- sqrt((-7)^2-4*1*(-10))/(2*1))
= 7/2-1/2*sqrt(89)

Sep 28, 2015

This cannot be factored using whole numbers. (It can be factored using some square roots.)

Explanation:

If we are to factor q^2-7q-10 using whole numbers, we would need to have (q +- "some wholenumber")(q +- "some other number")

Because we would use FOIL to multiply these two expression, we would need the two numbers (they will be the Last) to have a product of -10.

the Outside + Inside need to be -7q, so we needthe larger of the 'lasts' to be negative and the smaller to be positive.

The only possibilities are:

(q+1)(q-10) (which does not have -7q in the middle -- it has -9q)

(q+2)(q-5) (which does not have -7q in the middle -- it has -3q)

Since the only two ways to get q^2 First and -10# Last with a negative in the middle don't work. Nothing will work.

If you have learned how to solve q^2-7q-10 = 0 using square roots, you can factor the expression using those solutions.

The solutions are (7+sqrt89)/2 and (7-sqrt89)/2 and the factors are:

(x-(7+sqrt89)/2)(x-(7-sqrt89)/2)