Calculus with Applications (10th Edition)

$(3x + 5)(x - 2)$
Factor $3x^{2}-x-10$ First set up the x's in the binomial. $(cx +or- n)(cx +or- n)$ To have $3x^{2}$ in the original equation the two $x$'s in the binomial multiply together. Thus the $x$'s have coefficients. 3 is a prime number so its only factors are 3 and 1. Therefore in one of the parentheses the $x$ has a coefficient of 3 and the other has a coefficient of 1 $(3x +or- n)(1x +or- n)$ Now we need to find n in the expression above both n values must multiply to -10 and add (when the expression is expanded) -1x. Also we must take into account the 3 coefficient. To start list out the factors of -10: -1,10 1,-10 -2,5 2,5 Think which of these factor pairs when one number is multiplied by three and subtracted by the other leaves us with -1 Immediately we can discount the 1,10 pairs because there would be too large of a difference, leaving us to choose between the -2,5 and the 2,-5 factor pairs. $2*3=6$ and $5-6=1$ so we know that the $2$ will not be in the same parentheses as the $3x$ $(3x +or- 5)(1x +or- 2)$ Last we need to figure out the signs within the parentheses. As mentioned before $5-6=1$ so we want 6 to be negative, thus we write $-2$ $(3x +or- 5)(1x - 2)$ And because we want 5 to be positive we write $+5$ $(3x + 5)(x - 2)$ To check expand the expression and will match the original equation