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# factor completely the given polynomial 18x^2+51x+36

factor completely the given polynomial 18x^2+51x+36

There is the universal formula to factorize the quadratic expression: ax 2+ bx + c = a(x - x1)(x - x2) , where x1 and x2 are roots of quadratic equation ax2+ bx + c = 0 . To find the roots we can use the formula x = (-b ± √D)/2a , where D = b2 - 4ac . There are 2 roots if D>0 , there is 1root if D=0 , and zero roots in set of real numbers if D is negative.
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Let's start by taking 3 (common factor of 18, 51 and 36) out of parentheses 18x2 + 51x + 36 = 3(6x2+ 17x + 12) .... (1) and find the roots of equation 6x2+ 17x + 12 = 0 .... (2) ------> D = 172 - 4 · 6 · 12 = 1
x1,2 = (-17 ± √1)/(2 · 6) = (-17 ± 1)/12 -----> x1 = (-17 + 1)/12 = - 16/12 = -4/3
x2 = (-17 - 1)/12 = - 18/12 = -3/2
6x2+ 17x + 12 = 6(x -(-4/3))(x - (-3/2)) = (3x + 4)(2x + 3)
The final answer will be 3(3x + 4)(2x + 3)

First, it looks like we can pull a 3 out of each term:

3 (6x2 + 17x + 12)

now we need to examine the factors of the coefficents:

6x2 = 1x and 6x  or 2x and 3 x
12 = 1 and 12, 2 and 6, 3 and 4

Since we need the middle term to be odd, 3(2x + 3)(3x + 4) looks like our best combination.

Check by FOIL:

3(2x + 3)(3x + 4) = 3(6x2 + 8x + 9x + 12) = 3 (6x2 + 17x + 12) = 18x2 + 51x + 36   checked!