MATH SOLVE

3 months ago

Q:
# Factor the following problems SHOW YOUR WORK1. x^2+16x+242. x^2-10a+25

Accepted Solution

A:

x^2+16x+24: Think: What are the integer factors of 24? I'd list 1, 2, 3, 4, 6, 8, 12 and 24. Next: Look for a pair of factors from that list that:

(1)ADD up to 16 AND (2) whose product is 24. Unfortunately there seems to be no such pair from the above list.

So: Use the quadratic formula to find the roots of x^2+16x+24:

-16 plus or minus sqrt[ 256-4(1)(24) ]

x = --------------------------------------------------

2

-16 plus or minus sqrt(160) -16 plus or minus 4 sqrt(10)

= --------------------------------------- = ----------------------------------------

2 2

2(-8 plus or minus 2 sqrt(10) )

= ------------------------------------------- = -8 plus or minus 2sqrt(10)

2

If c is a root, (x-c) is a factor. Thus, if -8 plus 2sqrt(10) is a root,

(x+8-2sqrt(10) is a factor. Can you find the other factor?

x^2-10a+25 is much easier to factor. The sqrt of 25 is 5, so try x= -5 as a root. Dividing x-5 into x^2-10a+25, we get x-5. Therefore, the factors are

(x-5) and (x-5).

(1)ADD up to 16 AND (2) whose product is 24. Unfortunately there seems to be no such pair from the above list.

So: Use the quadratic formula to find the roots of x^2+16x+24:

-16 plus or minus sqrt[ 256-4(1)(24) ]

x = --------------------------------------------------

2

-16 plus or minus sqrt(160) -16 plus or minus 4 sqrt(10)

= --------------------------------------- = ----------------------------------------

2 2

2(-8 plus or minus 2 sqrt(10) )

= ------------------------------------------- = -8 plus or minus 2sqrt(10)

2

If c is a root, (x-c) is a factor. Thus, if -8 plus 2sqrt(10) is a root,

(x+8-2sqrt(10) is a factor. Can you find the other factor?

x^2-10a+25 is much easier to factor. The sqrt of 25 is 5, so try x= -5 as a root. Dividing x-5 into x^2-10a+25, we get x-5. Therefore, the factors are

(x-5) and (x-5).