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Question 549716: how do you factor the polynomial 2x squared plus 23x plus 45
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 2x^2 + 23x + 45 = (2x + 5) * (x + 9)
the factors are derived from:
9 * 5 = 45
2 * 9 + 1 * 5 = 18 + 5 = 23
the factors for 2x^2 had to be 2x and x.
that's the only way you can get 2x^2 because 2x * x = x^2
the factors for 45 could have been 45 * 1 and 5 * 9. not sure if there's other possible combinations but i tried those first.
i tried 45 * 1 and came up empty because:
2 * 45 = 90 and 1 * 1 = 1 for a total of 91.
i then tried 2 * 1 and 1 * 45 for a total of 46.
neither one was good.
i then tried 2 * 5 and 1 * 9 for a total of 10 + 9 = 19 - still no cigars.
i then tried 2 * 9 and 1 * 5 for a total of 18 + 5 = 23 - bingo !!!!
my factors then because (2x + 5) * (x + 9)
when you multiply these factors together using the basic concepts of distributive multiplication, then you get:
(2x + 5) * (x + 9) equals:
2x * (x + 9) + 5 * (x + 9) which becomes:
2x^2 + 18x + 5x + 45.
when you combine like terms, you get:
2x^2 + 23x + 45 which is the original quadratic equation.
with factoring of quadratic equations, you have to play around with the possible combinations.
your possible combinations for the coefficients of the x terms were 2 and 1.
your possible combinations for the constant terms were 45 and 1, 5 and 9.
it's then a matter of picking the right combinations that will work well with each other.
the right combination was 5 and 9 for the constants coupled with 2 and 1 for the coefficients of the x term.
the possible combinations were:
2 * 9 = 18
1 * 5 = 5
and:
2 * 5 = 10
1 * 9 = 9
when you added the results of those multiplications together, the right combination became:
2 * 9 = 18
1 * 5 = 5
for a total of 23.
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