Question

2x^2-7x-39 factorise

Answer 1

$P(x)=2x^2-7x-39\\\\factors, if\ rational\ factors\ exist\ are\ from: + or - \frac{factor\ of\ 39}{factor\ of\ 2}\\\\ie.,\ \ +\ or\ -\ \frac{1, 3, 13, 39}{1, 2},\\ie, 1,3,13,39,-1,-3,-13,-39,1/2,3/3,13/2,39/2,\\.\ \ \ \ -1/2,-3/2,-13/2,-39/2\\\\It\ is\ possible\ that\ there\ is\ no\ rational\ root.\\\\trying\ some\ values\ we\ find\ that\ P(-3)=0, => (x+3) is\ a\ factor.\\\\Then [x - (-39/(2*-3)) ] is another factor. \\\\P(x)=(x+3)2(x-13/2)=(x+3)(2x-13)$

====================  another way

find solutions / roots of quadratic expression P(x) = 2x^2-7x-39 =0

$2x^2-7x-39 =0\\\\roots=(7+-\sqrt{7^2+2*4*39})/4=(7+-19)/4=-3 or +13/2\\\\Hence (x+3)\ and\ (x-13/2)\ are \ factors.$

Answer 2

Answer:

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Step-by-step explanation: