you have to know that the factors will be
(2x-?)(x+?) where the ?'s must be divisors of 10.
So, the only possible factors are
(2x+10)(x-1)
(2x-10)(x+1)
(2x-5)(x+2)
(2x+5)(x-2)
(2x+2)(x-5)
(2x-2)(x+5)
(2x-1)(x+10)
(2x+1)(x-10)
With some practice, you can soon learn to find the factors without too much effort.
Of course, you can always rely on the quadratic formula. It says that
x = (1±√(1+80))/(2*2)
= (1±9)/4
= 10/4 or -8/4
= 5/2 or -2
So, you have factors of
(2x-5)(x+2)
I checked the answer in my textbook on what the answer is if I factor 2x^2-x-10
The answer is (2x-5)(x+2)
That answer makes sense to me if I expand it I'll get 2x^2-x-10.
My issue is I don't know how I would get 2x^2-x-10 in factored form
2 answers
There are several algorithms:
1. ----- This method seems to be taught these days:
for ax^2 + bx + c vs. 2x^2 - x - 10
- multiply a times c -----> 2(-10) = -20
- now look for factors of -20 which will add up to b
our b = -1
so a few quick tries shows (-5)(+4) = -20 and -5+4 = -1
- replace the middle term -x with -5x and +4x, the -5 and 4 coming from above
2x^2-x-10
= 2x^2 - 5x + 4x - 10
you now WILL BE ABLE to factor this by grouping
= x(2x - 5) + 2(2x - 5)
= (2x-5)(x+2)
2. ---- use the quadratic formula to find the roots, and form the factors from that.
2x^2 - x - 10 = 0
x = (1 ± √81)/4 = (1 ± 9)/4 = 5/2 or -2
if x= -2 , the factor is (x+2)
if x = 5/2 , the factor is (2x - 5)
and you got your factors!
btw, I often find b^2 - 4ac before attempting to find factors. If b^2 - 4ac is NOT a perfect square you will not be able to find any rational factors
3. ---- the way I do it:
- write down the factors of a and c in column form ignoring any ± signs
1 2 ------- 1 2
2 1 ------10 5
Now look for crossproduct that have a sum of the middle term coefficients if c > 0 , or look for crossproducts that have a difference of the middle term coefficient if c < 0
in our case c is negative, so notice that
2(2) - 1(5) = -1 <---- our middle term
You know your brackets will be (2x .....)(x .....)
the -5 and +4 can only go this way:
(2x - 5)(x + 4 )
This method is the fastest way, but requires some practise, and some number and common sense.
1. ----- This method seems to be taught these days:
for ax^2 + bx + c vs. 2x^2 - x - 10
- multiply a times c -----> 2(-10) = -20
- now look for factors of -20 which will add up to b
our b = -1
so a few quick tries shows (-5)(+4) = -20 and -5+4 = -1
- replace the middle term -x with -5x and +4x, the -5 and 4 coming from above
2x^2-x-10
= 2x^2 - 5x + 4x - 10
you now WILL BE ABLE to factor this by grouping
= x(2x - 5) + 2(2x - 5)
= (2x-5)(x+2)
2. ---- use the quadratic formula to find the roots, and form the factors from that.
2x^2 - x - 10 = 0
x = (1 ± √81)/4 = (1 ± 9)/4 = 5/2 or -2
if x= -2 , the factor is (x+2)
if x = 5/2 , the factor is (2x - 5)
and you got your factors!
btw, I often find b^2 - 4ac before attempting to find factors. If b^2 - 4ac is NOT a perfect square you will not be able to find any rational factors
3. ---- the way I do it:
- write down the factors of a and c in column form ignoring any ± signs
1 2 ------- 1 2
2 1 ------10 5
Now look for crossproduct that have a sum of the middle term coefficients if c > 0 , or look for crossproducts that have a difference of the middle term coefficient if c < 0
in our case c is negative, so notice that
2(2) - 1(5) = -1 <---- our middle term
You know your brackets will be (2x .....)(x .....)
the -5 and +4 can only go this way:
(2x - 5)(x + 4 )
This method is the fastest way, but requires some practise, and some number and common sense.
5 answers