How do I solve this Quadratic Equation: 2x^2 + 5x - 3 (by factoring)?

Question

I know how to factor Quadratic Equations that do not have a coefficient in the first part. The 2x^2 is what's messing me up. A friend told me you multiply the first coeffecient by the last constant (2x3 = 6) and find two terms that multiply to 6 and add up to 5. It ends up wrong.

Answer 1

You start by factoring the 2x^2:

(2x + .....) (x + .....)

Then you factor the -3 into -1 and 3, and put it in the parentheses:

(2x - 1) (x + 3)

Now multiply it out to check:

2x^2 -x +6x - 3 = 2x^2 + 5x - 3

So, the answer is (2x - 1) (x + 3)

Answer 2

Your friend was right: 2*3 = 6, just like 1*6 => 1x, 6x (6x-x). 6 determined the product of your potential coefficients. And pointed you to the fact that, since we have a negative coefficient involved, we cannot factor to a sum, but we will have to go for a difference (we will need two positive coefficients and two negative ones, in order to match everything properly).

Here we go:

2x^2 + 5x - 3 = 0 (5x can be also 6x-x)
2x^2 + 6x - x - 3 = 0
2x(x+3) - (x+3) = 0
(x+3)(2x-1)=0

x+3 = 0
x = -3

2x-1 = 0
2x= 1 /:2
x = 1/2

Check:

2x^2 + 5x - 3 = 0
x = -3
2(-3)^2 + 5(-3) - 3 = 0
18-15-3 = 0
18-18 = 0
OK

2x^2 + 5x - 3 = 0
x = 1/2
2(1/2)^2 + 5*1/2 - 3 = 0
2*1/4 + 5/2 - 3 = 0 (3 = 6/2)
1/2 + 5/2 - 6/2 = 0
6/2 - 6/2 = 0
0 = 0
OK

Answer 3

What you have to realize is that now instead of factoring x^2 into x and x, you're now factoring 2x^2 into 2x and x.

If you look at your outside terms, 2x^2 only has 2 possible factors (2x and x), and 3 only has 2 possible factors (3 and 1). Because the 3 is negative, you have to be subtracting either the 3 or the 1 (negative * positive = negative).

Because one of the terms is 2x, either the 3 or the 1 will be multiplied by 2 in the adding portion (middle term).

So really, what you're doing is finding 2 numbers a and b, where 2a + b = 5 and a*b = -3. Therefore, a = 3, b = -1.

Your factorization would then be (2x + 3)(x - 1).

Answer 4

2x^2 + 5x - 3
multiplying 2 X 3 = 6
as the two coefficients are (2) & (-3) and the product is minus (or-6)
Think of two numbers which when multiplied gives 6 and when substracted gives the cofficient of the middle term that is 5.
(if the product was + 6 we would have looked for addition giving 5 not subtraction.)
The two numbers are +6 & 1 and 5x = +6x-1x
now the question is
2x^2 + 5x - 3
or 2x^2 + 6x -1x - 3 = 2x (x+3) -1 (x+3 ) = (2x -1) (x+3)

Answer 5

The basic method of doing questions that have a 2x^2 (or anything other than a simple x^2 in them) is to plug and test.

One of the things you're definitely going to do is split up 2x^2 into 2x and x.

(2x + ?) (x + ?)

What goes in the question marks are factors of the third value, -3. What two numbers multiply to make -3? Let's test 3 and -1.

(2x + 3) (2x - 1)

At this point, what you want to test "The outside plus the inside", to see if you get the middle term. In this case, the outside is (-1)(2x) = -2x, and the inside is (3)(2x) = 6x.

Added together, -2x + 6x = 4x, which is NOT equal to the middle term.

Because those two didn't work, we try again, with other factors of -3. Let's try +1 and -3.

(2x + 1) (x - 3)

The outside (-6x) plus the inside (x) = -5x. Is this equal to the middle term? ALMOST; the only thing different is the sign. This should make a light bulb flash in your head that all you have to do is switch signs; that is, instead of +1 and -3, choose -1 and +3.

(2x - 1) (x + 3)

Test the outside (6x) plus the inside (-x), which equals 5x. This is our middle term, so that's the correct factorization.

That's the basic method of solving these. If you had something like this though:

4x^2 + 2x + 3, you're either going to have
(2x + ?) (2x + ?)
or
(4x + ?) (x + ?)

resulting in more things to test. Either way though, this is the basic "trial and error" method.

Answer 6

Answer coming up - it takes time to type

You write out the equation 2x^2 + 5x -3 = 0

In other words you write it in the order some squareds + or - some x's + - some numbers = 0

You are trying to put the answer as 2 brackets mutiplied together.





(x + 1)(x-2) = 0..........you have already learnt how to multiply this out

You got the answer to the x squareds by multiplying together the two x terms - in this case x times x = x^2

(x )(2x )...........is the only way you can get 2x^2

The only way you can get -3 is by multiply -3 by 1 or -1 by 3

Since you want to finish up with 5x, it looks promising to put the 3 where it will multiply by the 2x and the 1 where it will multiply by the x

(x......3)(2x.....1)

3 times 2x = 6x and 1 times x = x

As the number term is negative, you're looking for a difference (answer to a subtraction sum)

The 5x is positive, so the larger number is positive.

(x + 3)(2x - 1) = 0

check by multiply out......It works!

Now you can say x+3 = 0 , if so x = -3

or 2x -1 = 0, if so x = 1/2

Hope this helps.

This method is really doing "multiplying out" backwards - experimentally - seeing which values work.

You have to copy questions carefully as they don't all "work"

Good luck

Answer 7

2x^2 + 5x - 3 = 0

Start by setting up your two factors
and simplifying the squared term

(2x )(x ) = 0

one of the terms has to be 2x and the other x.

Now it's just a matter of plugging in the other parts.
-3 is acheived only by multiplying 3 x -1 or -3 x 1, so the possibilities are: (remember FOIL)

(2x - 3)(x + 1) = 0 2x^2 + 2x - 3x -3 = 0 2x^2 - x - 3 = 0 NOPE

(2x + 3)(x - 1) = 0 2x^2 - 2x + 3x -3 = 0 2x^2 + x - 3 = 0 NOPE

(2x + 1)(x - 3) = 0 2x^2 - 6x + x -3 = 0 2x^2 - 5x - 3 = 0 NOPE

(2x - 1)(x + 3) = 0 2x^2 + 6x - x -3 = 0 2x^2 + 5x - 3 = 0 That's the ticket!!

HTH

Answer 8

Its very simple!

2x^2 + 5x - 3 = 0

You take the FIRST TERM, that is (2x^2) and the LAST TERM that is (3)

and NEXT to your question, do this:

2x 3
1x 1
(factors od 2x^2!) (factors of 3!)


you must find FACTORS of each term, do you see how i've done it?

2x^2 can be split into 2x and 1x, as 2x multiplied by 1x = 2x^2
3 can be split into 3 and 1, as 3 multplied by 1 = 3

now, you CROSS MULTIPLY!

that is, you draw a cross over your sum,
and multiply 2x by 3 (you see how ive crossed?)
and multiply 1x by 1
write the answer at the next to the number on the right!

your sum will look like this

2x 1 1x
1x 3 6x


now, you want to get a POSITIVE 5!
so you must MINUS OR ADD 1 and 6 to get POSITVE 5

so you say NEGATIVE 1 plus POSITIVE 6 is going to give you POSITIVE 5!

so right THOSE SIGNS next to your 1 and 6 in your answer
it should look like this:

2x 1 -1x
1x 3 +6x
= +5x

now, take those SIGNS, and move them between your factors!

2x - 1
1x + 3

NOW YOU HAVE FACTORISED!!

the answer therefore is

2x^2 + 5x -3
= (2x-1)(x+3)

therefore,

2x=1 or x= -3
x=1/2


ITS MUCH SIMPLIER ONCE YOUVE MASTERED THE METHOD!! T TAKES ABOUT 5-10 SECONDS WHEN YOU CAN DO THIS METHOD PROPERLY, SO DONT BE DISCOURAGED!

Answer 9

2x^2+5x-3=0

the answer is (2x-1)(x+3)=0

2x-1=0 x+3=0
x=1/2 and x= -3

identify the coefficient ,then,factorise it
e.g= 2x^2 can be factorised as 2x(x)
therefore,(2x )(x )=0

then,you factorise 3 into this --> 1(3)
therefore,(2x 1)(x 3)=0

all you have to do is to put the sign which is match with the equation

(2x-1)(x+3)=0

then,you need to recheck back by expanding the equation

Answer 10

~** Grouping Number Method for
Factoring Trinomials of the Form ax² + bx + c **~

2x² + 5 - 3

1) Obtain the "grouping number" (a)(c).
The grouping number is (a)(c) = (2)(-3) = (-6)

2) Find the factor pair of the grouping number whose sum is b.
The factor pairs of (-6) are as follows:
(6)(-1) ... (6) + (-1) = 5
(-6)(1) ... (-6) + (1) = -5
(3)(-2) ... (3) + (-2) = 1
(-3)(2) ... (-3) + (2) = -1
Since b = 5, we want the factor pair of (-6) whose sum is 5. Therefore, we select the factors (6) and (-1).

3) Use those two factors to write bx as the sum of two terms.
We use the numbers (6) and (-1) to write 5x as the sum of (6x) and (-1x).
2x² + 5x - 3 = 2x² + 6x - 1x - 3

4) Factor by grouping.
2x² + 6x - 1x - 3 --- Rearange terms and add parentheses for simplicity...
(2x² - 1x) + (6x - 3) --- Factor out x from the fisrt set, and 3 from the second set...
x(2x - 1) + 3(2x - 1) --- Factor out (2x -1)...
(2x - 1)(x + 3)

ANSWER: 2x² + 5x - 3 = (2x - 1)(x + 3)

CHECK: Check using the F.O.I.L. method.
(2x - 1)(x + 3)
= (2x)(x) + (2x)(3) + (-1)(x) + (-1)(3)
= 2x² + 6x + (-1x) + (-3)
= 2x² + 5x - 3 --- Answer checks out!