Factor completely: 2x2 + 6x − 80 (2x − 5)(x + 8) (2x − 10)(x + 8) 2(x − 5)(x + 8) 2(x − 10)(x + 8)
the first question is 8 and the second question is 5 since 8(-5)=-40 and 8+(-5)=3
\[\large 2x^2 + 6x - 80\] \[\large 2(x^2 + 3x - 40)\] \[\large 2(x^2 + 8x-5x - 40)\] \[\large 2(x(x + 8)-5(x +8))\] \[\large 2(x-5)(x + 8)\] So the answer is choice C
one day i am going to learn this factor by grouping thing
satellite i thought i taught you :(
yes you did. you taught me, but i didn't learn it. (you been there before i am sure). care to try again?
\[ax^2+bx+c\] to factor something of this form: Find factors of a*c that have product a*c and have sum b
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
that is the same thing as "factoring" without grouping.
factor by grouping is how i did that one proof on the quadratic formula
factor by grouping is the coolest!
oooh right. like completing the square. or that great sophie germain trick
i remember that trick a bit
oh jim hasn't seen my proof on the quad formula yet : 0
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)
i like jim's explanations
the pdf isn't showing up for me...
like when you click it, it doesn't show anything?
apparently it doesn't like firefox, but chrome works fine, weird...
well completing the square is still a better way to prove the quadratic formula but this works too
i am a big fan of completing the square in all circumstances except finding a vertex (lol)
for example in factoring \[x^4+x^2+1\]
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