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anonymous
 one year ago
Determine two different values of “b” in x2 + bx + 30 so that the expression can be factored into the product of two binomials. Explain how you determined those values, and show each factorization. Explain how your process would change if the expression was 2x2 + bx + 30.
anonymous
 one year ago
Determine two different values of “b” in x2 + bx + 30 so that the expression can be factored into the product of two binomials. Explain how you determined those values, and show each factorization. Explain how your process would change if the expression was 2x2 + bx + 30.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So like how would I do this if b = 5 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The polynomial is not factorable with rational numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0b = 5 is not possible....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0b could be 17 or 13. If 17 then you have: (x+15)(x+2)=x2+15x+2x+30 If 13 then you have: (x+10)(x+3)=x2+10x+3x+30

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im confused.. which one is right ? im here because I really have no idea what to do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the question asked for what two numbers could b be.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02x2 is what and =30 is what imma let that guy take over

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you know it must be in the form (x+c)(x+d), so you expand this to x^2+(c+d)x+cd, equate the coefficients with the equation you had originally so c+d=b and cd=30 by choosing an arbitrary value for c you can work out d, like so c=2, 2*d=30, so d=15, now you can work out b, 2+15=b so b=17. you can do this for as many values for c or d as you want. for 2x^2+bx+30 I would do the same except I would expand (2x+c)(x+d) instead, you would then do the same as before, equating coefficients and choosing arbitrary values for c or d.
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