Mathematics 11 Online
OpenStudy (anonymous):

Factor completely: 2x2 + 6x − 80 (2x − 5)(x + 8) (2x − 10)(x + 8) 2(x − 5)(x + 8) 2(x − 10)(x + 8)

myininaya (myininaya):

\[2x^2+6x-80=2(x^2+3x-40)=2(x+?)(x-?)\]

myininaya (myininaya):

the first question is 8 and the second question is 5 since 8(-5)=-40 and 8+(-5)=3

jimthompson5910 (jim_thompson5910):

\[\large 2x^2 + 6x - 80\] \[\large 2(x^2 + 3x - 40)\] \[\large 2(x^2 + 8x-5x - 40)\] \[\large 2(x(x + 8)-5(x +8))\] \[\large 2(x-5)(x + 8)\] So the answer is choice C

OpenStudy (anonymous):

one day i am going to learn this factor by grouping thing

myininaya (myininaya):

satellite i thought i taught you :(

OpenStudy (anonymous):

yes you did. you taught me, but i didn't learn it. (you been there before i am sure). care to try again?

myininaya (myininaya):

\[ax^2+bx+c\] to factor something of this form: Find factors of a*c that have product a*c and have sum b

OpenStudy (anonymous):

we got \[x^2+3x-40\] and please don't say "two numbers whose product is -40 and whose sum is 3" because that is how i do it

myininaya (myininaya):

lol

OpenStudy (anonymous):

that is the same thing as "factoring" without grouping.

myininaya (myininaya):

factor by grouping is how i did that one proof on the quadratic formula

myininaya (myininaya):

factor by grouping is the coolest!

OpenStudy (anonymous):

oooh right. like completing the square. or that great sophie germain trick

myininaya (myininaya):

i remember that trick a bit

myininaya (myininaya):

oh jim hasn't seen my proof on the quad formula yet : 0

myininaya (myininaya):

jimthompson5910 (jim_thompson5910):

Example: Factor 9x^2-6x-8 Step 1) Multiply the first and last coefficients: 9*(-8)=-72 Step 2) Find two numbers that both multiply to -72 and add to -6. Those numbers are -12 and 6 Step 3) Replace -6x with -12x+6x (notice how -12x+6x adds back to -6x). The -12x and -6x are drawn exactly from those two numbers we found in step 2). So 9x^2-6x-8 turns into 9x^2-12x+6x-8 Step 4) Now factor the expression by grouping 9x^2-12x+6x-8 (9x^2-12x)+(6x-8) ... pair off terms in 2 groups (how you pair depends on the problem) 3x(3x-4)+(6x-8) ... factor out the GCF 3x from the first group 3x(3x-4)+2(3x-4) ... factor out the GCF 2 from the second group Notice we now have a common factor of 3x-4, factor this out to get (3x+2)(3x-4) So 9x^2-6x-8 = (3x+2)(3x-4)

myininaya (myininaya):

i like jim's explanations

myininaya (myininaya):

so perfect

jimthompson5910 (jim_thompson5910):

the pdf isn't showing up for me...

myininaya (myininaya):

myininaya (myininaya):

like when you click it, it doesn't show anything?

jimthompson5910 (jim_thompson5910):

apparently it doesn't like firefox, but chrome works fine, weird...

myininaya (myininaya):

well completing the square is still a better way to prove the quadratic formula but this works too

OpenStudy (anonymous):

i am a big fan of completing the square in all circumstances except finding a vertex (lol)

OpenStudy (anonymous):

for example in factoring \[x^4+x^2+1\]