How do you factor q^2 - 7q - 10q2−7q−10?

2 Answers
Farzana H.
Sep 28, 2015

x_1= 7/2+1/2*(sqrt(89))x1=72+12⋅(√89)
x_2= 7/2-1/2*(sqrt(89))x2=72−12⋅(√89)

Explanation:

quadratic formula
x_1,_2 =(-b+- sqrt(b^2-4ac))/(2a)x1,2=−b±√b2−4ac2a
here a=1,b=-7,c=-10a=1,b=−7,c=−10
so
x_1 =(-(-7)+ sqrt((-7)^2-4*1*(-10))/(2*1))x1=⎛⎜ ⎜⎝−(−7)+√(−7)2−4⋅1⋅(−10)2⋅1⎞⎟ ⎟⎠
= 7/2+1/2*sqrt(89)=72+12⋅√89
x_2 =(-(-7)- sqrt((-7)^2-4*1*(-10))/(2*1))x2=⎛⎜ ⎜⎝−(−7)−√(−7)2−4⋅1⋅(−10)2⋅1⎞⎟ ⎟⎠
= 7/2-1/2*sqrt(89)=72−12⋅√89

Jim H
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-10q2−7q−10 using whole numbers, we would need to have (q +- "some wholenumber")(q +- "some other number")(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 -10−10.

the Outside + Inside need to be -7q−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)(q+1)(q−10) (which does not have -7q−7q in the middle -- it has -9q−9q)

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

Since the only two ways to get q^2q2 First and 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 = 0q2−7q−10=0 using square roots, you can factor the expression using those solutions.

The solutions are (7+sqrt89)/27+√892 and (7-sqrt89)/27−√892 and the factors are:

(x-(7+sqrt89)/2)(x-(7-sqrt89)/2)(x−7+√892)(x−7−√892)

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