Question 524: How do I factor 6x^2-7x-4?
Answer by AnlytcPhil(1806)   (Show Source):
You can put this solution on YOUR website! How do I factor 6x^2-7x-4
`
Make a list of all the possible ways for
the FIRSTS to have product 6x²
`
(1x ` )(6x ` )
(2x ` )(3x ` )
`
Make a list of all the ways for the
LASTS to have product 4
`
( ` 1)( ` 4)
( ` 2)( ` 2)
`
Pick one of the above lists to extend to
include all the "reversals", i.e., the
"reversal of "1×4" is "4×1"
`
I pick the list of factors of 4, because
all I have to do is add on the reversal
"4×1". So we have these lists of
possibilities for the FIRSTS:
`
(1x ` )(6x ` )
(2x ` )(3x ` )
`
and these possibilities for lasts:
`
( ` ` 1)( ` ` 4)
( ` ` 2)( ` ` 2)
( ` ` 4)( ` ` 1)
`
Now put each of the first two with each of
the last three
`
(1x ` 1)(6x ` 4)
(1x ` 2)(6x ` 2)
(1x ` 4)(6x ` 1)
(2x ` 1)(3x ` 4)
(2x ` 2)(3x ` 2)
(2x ` 4)(3x ` 1)
`
Strike through any of these which contains a
parenthetical binomial from which a common
factor can be taken out
`
We strike through (eliminate) the first one
because 2 can be taken out of (6x ` 4)
`
We strike through (eliminate) the second one
because 2 can be taken out of (6x ` 2)
`
We strike through (eliminate) the fifth one
because 2 can be taken out of (2x ` 2)
`
We strike through (eliminate) the sixth one
because 2 can be taken out of (2x ` 4)
`
So now our list is
`
(1x ` 1)(6x ` 4)
(1x ` 2)(6x ` 2)
(1x ` 4)(6x ` 1)
(2x ` 1)(3x ` 4)
(2x ` 2)(3x ` 2)
(2x ` 4)(3x ` 1)
`
So all we have left are these two.
`
(1x ` 4)(6x ` 1)
(2x ` 1)(3x ` 4)
`
Since the last term of the trinomial
6x^2-7x-4 is negative, the signs are different,
so we need only try
`
(1x + 4)(6x - 1)
(2x + 1)(3x - 4)
(1x - 4)(6x + 1)
(2x - 1)(3x + 4)
`
Multiplying them out
`
(1x + 4)(6x - 1) = 6x² + 23x - 4
(2x + 1)(3x - 4) = 6x² - 5x - 4
(1x - 4)(6x + 1) = 6x² - 23x - 4
(2x - 1)(3x + 4) = 6x² + 5x - 4
`
None of those are 6x²-7x-4, so 6x²-7x-4 is
prime and thus will not factor.
`
Edwin
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