Can someone help me figure this out? Select one of the factors of x^2 – 17x + 72 I will leave the possible answers here as reference: a)x - 8 b)x - 2 c)x + 9 d)x + 12 I am taking Algebra 2 and am currently pulling a high "B". I want to bring that up to an "A". Can anyone explain the steps I need to figure this out? :D
(x-8)(x-9)
How did you figure that out?
You should two numbers whose product is 72 And whose sum is 17
*find
x^2-17x+72=0 x^2-8x-9x+72=0 x(x-8)-9(x-8)=0 (x-8)(x-9)=0
Alright, that makes sense. Thanks! I just got the same answer! Thank you so much!
Ok, so I have this problem now which is 4x^2 - 8x - 140. A) x + 5 B) 8 C) x - 4 D) x - 2 I tried getting the answer by finding two numbers that get the product of -140 and the sum of -8. I ended up with (x + 5)(x - 13) so that would give me -8 but not a product of -140. I am hopelessly confused, can anyone clear this up for me?
You can simplify that by taking out the common factor "4"
So, it would end up being 4(x + y)(x + y)
y being a placeholder
Divide through out by 4 you get x^2-2x-35=0 x^2-7x+5x-35=0 X(x-7)-5(x-7)=0 (x-7)(x-5)=0
yeah but first find what the quad equation would look like
*x+5
Factoring equations sucks.. :(
scasm already did the work for you though
I see that, I was trying to figure it out so I didn't have to keep asking for help. :P
you'll eventually get used to it :)) its really frustrating at first you know
Alright, I think this will be the last question I need to understand it. 2x^2 + 16x - 18 I factored out the common factor which is 2. so now I have x^2 + 8x - 9 I imagine the quad would be x^2 + 4x + 4x - 9. what can I do from here or what did I do wrong?
You dont get 4 and 4. Since the last number is 9, u have this|dw:1334685833822:dw| with some permutations and combinations get two numbers such that their product is 9 and their sum is 8. Here 3*3*1 are factors of 9 .So its 9 and 1 equation will be x^2+9x-x-9=0 Here the product should be -9 as indicated by the equation, so the factors are 9 and -1 which also complies with their sum 9-1=8 which is right in the middle of the eqn
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