Question

help please

Answer 1

@satellite73 @texaschic101 @whpalmer4

Answer 2

@amistre64

Answer 3

i have a system of multipying first and last terms to find factors that add to the middle term

since 5*2 = 10, and 5+2 = 7, it seems that redundant; but then i set up the factoring as

(x+5/5) (x+2/5)

and simplify, if theres still a fraction left over pull the bottom out front

(x+1) (x+2/5)

(x+1) (5x+2)

and thats it

Answer 4

That's a clever approach, and one that I don't recall seeing. It took me a few minutes to figure out exactly what you were doing (because 5 was used in two different ways), so I'll do a couple of different examples for the benefit of anyone else who sees this (and practice for myself!)

\[2x^2+11x+12\]2*12 = 24, which has factors 1,2,3,4,6,8,12,24. A pair that adds to 11 is 3 and 8. We take the 2 from \(2x^2\) and use it as the denominator of our fractions, and use 3 and 8 for the numerators, giving us
\[(x+\frac{3}{2})(x+\frac{8}{2})\]A first simplification gives us\[(x+\frac{3}{2})(x+4)\]Now we multiply the first term by 2 to clear the fraction, giving us\[2(x+\frac{3}{2})(x+4) = (2x+3)(x+4)\]Quick check to make sure it is correct:
\[(2x+3)(x+4) = 2x*x+2x*4+3*x+3*4 \]\[= 2x^2+8x+3x+12 = 2x^2+11x+12\checkmark\]

\[4x^2+7x-2\]-2*4 = -8, which has factors 1,2,4,8 (one of the factors will have to be negative). -1 and 8 add to 7.
\[(x+\frac{-1}{4})(x+\frac{8}{4})\]First pass of simplification gives
\[(x-\frac{1}{4})(x+2)\]Multiply the first term by 4 to clear the fraction\[4(x-\frac{1}{4})(x+2)=(4x-1)(x+2)\]Checking\[(4x-1)(x+2) = 4x*x+4x*2-1*x-1*2 = 4x^2+7x-2\checkmark\]

Finally, a tricky case:
\[4x^2-4x-15\]4*-15 = -60, factors that add to -4 are 6 and -10.
\[(x+\frac{6}{4})(x+\frac{-10}{4})\]Nothing simplifies to an integer, but we can't go on yet or we'll get the wrong answer! We need to fully reduce those fractions.\[(x+\frac{3}{2})(x-\frac{5}{2})\]Now we can multiply each one by 2 to clear out the fraction:\[2(x+\frac{3}{2})*2(x-\frac{5}{2}) = (2x+3)(2x-5)\]Checking:\[(2x+3)(2x-5) = 2x*2x+2x*(-5)+3*2x+3*(-5) = 4x^2-4x-15\checkmark\]

Notice if we hadn't reduced those fractions, we would end up with\[(4x-10)(4x+6) = 16x^2+24x-40x-60 = 16x^2-16x-60 \]\[= 4(4x^2-4x-15)\]

Answer 5

:)