quadratic equation: I factor as 6x^2 + x - 2, but book says (2x - 1)(3x + 2)

[SusanCoutu]

On a quadratic equation such as 6x^2 + x - 2, I get these results as factors, (x - 1/2)(x + 2/3), from using quadratic formula. Yet my answer book specifies these factors, (2x - 1)(3x + 2), as the answer in final form. I can see why both are equivalent upon performing the specified operations, however the latter produces a result strictly in integers. What I want to know is how exactly am I to transform my answer into the standard form they want? I can't see how I could multiply the first factor, (x - 1/2), by 2 only while also multiplying the second factor, (x + 2/3), by 3 only. Any help?

 

[Dr.Peterson]

On a quadratic equation such as 6x^2 + x - 2, I get these results as factors, (x - 1/2)(x + 2/3), from using quadratic formula. Yet my answer book specifies these factors, (2x - 1)(3x + 2), as the answer in final form. I can see why both are equivalent upon performing the specified operations, however the latter produces a result strictly in integers. What I want to know is how exactly am I to transform my answer into the standard form they want? I can't see how I could multiply the first factor, (x - 1/2), by 2 only while also multiplying the second factor, (x + 2/3), by 3 only. Any help?

Your factors, you should notice, do not produce the given expression, but rather (x - 1/2)(x + 2/3) = x^2 + 1/6 x - 1/3. So you have not yet factorized the given expression. But if you multiply that by 6, you get 6x^2 + x - 2, as desired. The factorization (2x - 1)(3x + 2) is not merely "the standard form they want", but the correct factorization over the integers.

What's happening here is that, given zeros p and q of ax^2 + bx + c, we have ax^2 + bx + c = a(x - p)(x - q). This is because (x - p)(x - q) has the right zeros, but so does any multiple of that. By multiplying by a, we get the right leading term, so that the original and the factored form are actually equal. Then you will always be able to find factors mn = a so that if we multiply (x-p) and (x-q) by m and n respectively, all coefficients are integers: mn(x - p)(x - q) = m(x - p) n(x - q) = (mx - mp)(nx - nq), where all coefficients are integers.

In your example, you just have to look at the denominators, 2 and 3, and recognize that their product is 6, which is your "a". So you multiply (x - 1/2) by 2 and (x + 2/3) by 3 to obtain the required factorization.

 

[Harry_the_cat]

On a quadratic equation such as 6x^2 + x - 2, I get these results as factors, (x - 1/2)(x + 2/3), from using quadratic formula. Yet my answer book specifies these factors, (2x - 1)(3x + 2), as the answer in final form. I can see why both are equivalent upon performing the specified operations, however the latter produces a result strictly in integers. What I want to know is how exactly am I to transform my answer into the standard form they want? I can't see how I could multiply the first factor, (x - 1/2), by 2 only while also multiplying the second factor, (x + 2/3), by 3 only. Any help?

Just a comment on your language: "a quadratic equation such as 6x^2 + x - 2". This is an expression not an equation (there is no = sign).​

 

[HallsofIvy]

On a quadratic equation such as 6x^2 + x - 2, I get these results as factors, (x - 1/2)(x + 2/3), from using quadratic formula.

Are you clear on what "factor" means? (x- 1/2)(x+ 2/3) are NOT factors of 6x^2+ x- 2 because multiplying them gives x^2+ x/6- 1/3 not 6x^2+ x- 2. You could write it as tree factors, 6(x- 1/2)(x+ 2/3).

Yet my answer book specifies these factors, (2x - 1)(3x + 2), as the answer in final form. I can see why both are equivalent upon performing the specified operations, however the latter produces a result strictly in integers. What I want to know is how exactly am I to transform my answer into the standard form they want? I can't see how I could multiply the first factor, (x - 1/2), by 2 only while also multiplying the second factor, (x + 2/3), by 3 only. Any help?

 

[Dr.Peterson]

Are you clear on what "factor" means? (x- 1/2)(x+ 2/3) are NOT factors of 6x^2+ x- 2 because multiplying them gives x^2+ x/6- 1/3 not 6x^2+ x- 2. You could write it as three factors, 6(x- 1/2)(x+ 2/3).

I'd rather say that (x-1/2) and (x+2/3) are factors (over the rational numbers), but (x-1/2)(x+2/3) does not constitute a factorization (that is, a way of writing it as a product of factors). In the same way, 2 and 3 are factors of 30; but 2*3 is not a factorization of 30.

But I agree that Susan's use of words, and perhaps misunderstanding of them, may be part of the problem.

 

[Steven G]

Are you clear on what "factor" means? (x- 1/2)(x+ 2/3) are NOT factors of 6x^2+ x- 2 because multiplying them gives x^2+ x/6- 1/3 not 6x^2+ x- 2. You could write it as three factors, 6(x- 1/2)(x+ 2/3).

Actually, (x- 1/2)(x+ 2/3) are NOT factors at all. (x- 1/2) AND (x+ 2/3) ARE factors of 6x^2+ x- 2. Yes, of course 6 is another factor of 6x^2+ x- 2. The three factors of 6x^2+ x- 2, in my opinion, are (is) not 6(x- 1/2)(x+ 2/3). The three factors of 6(x- 1/2)(x+ 2/3) are 6, (x- 1/2) and (x+ 2/3). Factors are separated by multiplication symbols.

 

[Deleted member 4993]

Actually, (x- 1/2)(x+ 2/3) are NOT factors at all. (x- 1/2) AND (x+ 2/3) ARE factors of 6x^2+ x- 2. Yes, of course 6 is another factor of 6x^2+ x- 2. The three factors of 6x^2+ x- 2, in my opinion, are (is) not 6(x- 1/2)(x+ 2/3). The three factors of 6(x- 1/2)(x+ 2/3) are 6, (x- 1/2) and (x+ 2/3). Factors are separated by multiplication symbols.

(x- 1/2)(x+ 2/3) is a factor of (6x^2+ x- 2) - just like 4 is a factor of 24.

 

[stapel]

What I want to know is how exactly am I to transform my answer into the standard form they want?

If you use a more standard method (that is, by doing the actual factorization process, rather than working backwards from the Quadratic Formula), you'll be more likely to obtain the expected forms. (Of course, you'll also need to check your work, so you note things like "x*x does not equal 6x^2, so maybe there's an error here".)

 

[JeffM]

On a quadratic equation such as 6x^2 + x - 2 ... What I want to know is how exactly am I to transform my answer into the standard form they want? ... Any help?

Leaving aside any errors that you made, let's try your method, which WILL work. Assuming a is not zero,

\(\displaystyle p = \dfrac{-\ b + \sqrt{b^2 - 4ac}}{2a} \text { and }q = \dfrac{-\ b - \sqrt{b^2 - 4ac}}{2a} \iff a(x - p)(x - q) = ax^2 + bx + c\).

So let's take your problem.

\(\displaystyle \sqrt{b^2 - 4ac} = \sqrt{1^2 - 4(6)(-\ 2)} = \sqrt{1 + 48} = \sqrt{49} = 7.\)

\(\displaystyle \therefore p = \dfrac{-\ 1 + 7}{2 * 6} = \dfrac{1}{2} \text { and } q = \dfrac{-\ 1 - 7}{12} = -\ \dfrac{2}{3} \implies -\ p = -\ \dfrac{1}{2} \text { and } -\ q = \dfrac{2}{3}.\)

\(\displaystyle \therefore 6 \left ( x - \dfrac{1}{2} \right ) \left ( x + \dfrac{2}{3} \right ) = 6x^2 + x - 2.\)

Now the product on the left is a perfectly valid factoring of the quadratic on the right. So in one sense, that is your answer. However, in this particular case, we can simplify the product on the left as follows:

\(\displaystyle 6 \left ( x - \dfrac{1}{2} \right ) \left ( x + \dfrac{2}{3} \right ) = 3 * 2 \left (x - \dfrac{1}{2} \right ) \left ( x + \dfrac{2}{3} \right ) =\)

\(\displaystyle (2x - 1) * 3 \left ( x + \dfrac{2}{3} \right ) = (2x - 1)(3x + 2).\)

Whether and how you distribute the 6 is arbitrary, but most people like to clear fractions if possible.

 

[Dr.Peterson]

\(\displaystyle \therefore 6 \left ( x - \dfrac{1}{2} \right ) \left ( x + \dfrac{2}{3} \right ) = 6x^2 + x - 2.\)

Now the product on the left is a perfectly valid factoring of the quadratic on the right. So in one sense, that is your answer. However, in this particular case, we can simplify the product on the left as follows:

\(\displaystyle 6 \left ( x - \dfrac{1}{2} \right ) \left ( x + \dfrac{2}{3} \right ) = 3 * 2 \left (x - \dfrac{1}{2} \right ) \left ( x + \dfrac{2}{3} \right ) =\)

\(\displaystyle (2x - 1) * 3 \left ( x + \dfrac{2}{3} \right ) = (2x - 1)(3x + 2).\)

Whether and how you distribute the 6 is arbitrary, but most people like to clear fractions if possible.

Here are two additional facts worth noting about this technique of "reverse engineering" a factorization from the zeros:

First, it will always be possible to clear fractions in this situation (namely, where we are trying to factor a quadratic with integer coefficients and rational zeros). We know this because we know it can be factored over the integers, and that the linear factors must be multiples of the rational factors you have formed, so there are linear factors with integer coefficients that we obtain by multiplying -- namely, by factors of a.

Second, we can get directly to the proper factoring over the integers with a slightly different approach. We have found the zeros 1/2 and -2/3; rather than using x - 1/2 and x + 2/3 as the factors, we can instead directly write 2x - 1 and 3x + 2, since if x - 1/2 = 0, then 2x - 1 = 0 by multiplying by 2. That is, for a rational zero m/n, we can write factor nx - m, which equals 0 when x = m/n. Therefore, we know that (2x - 1)(3x + 2) is a quadratic polynomial with the same zeros as 6x^2 + x - 2; and any quadratic polynomial with the same zeros and the same "a" is, in fact, the same polynomial. So this method will always work.

Finally, we should keep in mind that what is commonly just called "factoring" in elementary texts is actually "factoring over the integers", so that integer coefficients are expected; 6(x - 1/2)(x + 2/3) typically will not be accepted, even though in other contexts it is indeed a valid factorization. We don't know for sure the context of the OP.

 

[Steven G]

(x- 1/2)(x+ 2/3) is a factor of (6x^2+ x- 2) - just like 4 is a factor of 24.

I suspect you are correct. I just generally think of them as two factors instead of one.

 

[lookagain]

Actually, (x- 1/2)(x+ 2/3) [one quantity] are <--------

NOT factors at all. (x- 1/2) AND <-------

(x+ 2/3) ARE factors of 6x^2+ x- 2.

Jomo, look at a different aspect of this that might have helped clear up some of this. Look at the grammar.
(x - 1/2)(x + 2/3) is one product. It's one quantity. There needs to be the word "is" right after it instead of
"are" as you have it. That product is a factor, as well as 6 is a factor. In the second sentence above, the word
"and" is correct, because it separates two quantities.

 

[sinx]

On a quadratic equation such as 6x^2 + x - 2, I get these results as factors, (x - 1/2)(x + 2/3), from using quadratic formula. Yet my answer book specifies these factors, (2x - 1)(3x + 2), as the answer in final form. I can see why both are equivalent upon performing the specified operations, however the latter produces a result strictly in integers. What I want to know is how exactly am I to transform my answer into the standard form they want? I can't see how I could multiply the first factor, (x - 1/2), by 2 only while also multiplying the second factor, (x + 2/3), by 3 only. Any help?

because (x - 1/2)(x + 2/3) is not an equation.
if it were an equation, you could not multiply one factor by 2 and the other by 3.

however,
6x²+x-2=6(x - 1/2)(x + 2/3)
=2(x - 1/2)3(x + 2/3)

 

[JeffM]

Here are two additional facts worth noting about this technique of "reverse engineering" a factorization from the zeros:

First, it will always be possible to clear fractions in this situation (namely, where we are trying to factor a quadratic with integer coefficients and rational zeros). We know this because we know it can be factored over the integers, and that the linear factors must be multiples of the rational factors you have formed, so there are linear factors with integer coefficients that we obtain by multiplying -- namely, by factors of a.

Second, we can get directly to the proper factoring over the integers with a slightly different approach. We have found the zeros 1/2 and -2/3; rather than using x - 1/2 and x + 2/3 as the factors, we can instead directly write 2x - 1 and 3x + 2, since if x - 1/2 = 0, then 2x - 1 = 0 by multiplying by 2. That is, for a rational zero m/n, we can write factor nx - m, which equals 0 when x = m/n. Therefore, we know that (2x - 1)(3x + 2) is a quadratic polynomial with the same zeros as 6x^2 + x - 2; and any quadratic polynomial with the same zeros and the same "a" is, in fact, the same polynomial. So this method will always work.

Finally, we should keep in mind that what is commonly just called "factoring" in elementary texts is actually "factoring over the integers", so that integer coefficients are expected; 6(x - 1/2)(x + 2/3) typically will not be accepted, even though in other contexts it is indeed a valid factorization. We don't know for sure the context of the OP.

This was a very valuable addition to my post.