How to factor 4m^2 + 4m - 15

**shenton** · November 25th 2006, 07:59 PM

How to factor this:

4m^2 + 4m - 15

The answer key given is (2m + 5) (2m - 3).

Split middle term doesn't work here since I don't think I can find the sum/difference of 2 numbers that produce 4 for the product of 60.

What steps do I need to do to reach (2m + 5) (2m - 3)?

Thanks.

**Quick** · November 25th 2006, 08:13 PM

Originally Posted by shenton

How to factor this:

4m^2 + 4m - 15

The answer key given is (2m + 5) (2m - 3).

Split middle term doesn't work here since I don't think I can find the sum/difference of 2 numbers that produce 4 for the product of 60.

What steps do I need to do to reach (2m + 5) (2m - 3)?

Thanks.

I have a method to split the middle!

Given an expression: $ax^2+bx+c$

The middle terms are: $\frac{b\pm\sqrt{b^2-4ac}}{2}$

Note: it looks like the quadratic formula, but it doesn't have an "a" in the denominator and it's just "b" to start, not negative "b"

Substitute: $\frac{4\pm\sqrt{4^2-4(4)(-15)}}{2}$

Thus: $\frac{4}{2}\pm\frac{\sqrt{16+240}}{2}$

Therefore: $2\pm\frac{\sqrt{256}}{2}$

Then: $2\pm\frac{16}{2}$

Thus: $2\pm8$

So you can rewrite $4m^2 + 4m - 15$ into $4m^2-6m+10m-15$

So Factor: $2m(2m-3)+5(2m-3)$

Then add: $(2m+5)(2m-3)$

**Quick** · November 25th 2006, 08:19 PM

I would just like to say that my method is PROVEN (to all you skeptics out there) and was designed for excel, or if you're really stuck on a problem

**Soroban** · November 25th 2006, 08:26 PM

Hello, shenton!

How to factor this: . $4m^2 + 4m - 15$

Answer: . $(2m + 5) (2m - 3)$

I don't think I can find the sum/difference of 2 numbers that produce 4 for the product of 60.
. . . Sure you can!

Divide 60 by 1,2,3,4,... and keep the ones that "come out even".

$60 \div 1\:=\:60\quad\Rightarrow\quad 1\cdot60$
$60 \div 2 \:=\:30\quad\Rightarrow\quad 2\cdot30$
$60\div 3 \:=\:20\quad\Rightarrow\quad 3\cdot20$
$60 \div 4\:=\:15\quad\Rightarrow\quad 4\cdot15$
$60 \div 5\:=\:12\quad\Rightarrow\quad 5\cdot12$
$60 \div 6 \:=\:10\quad\Rightarrow\quad 6\cdot10\;\;\leftarrow\text{ a difference of 4 !}$

We want $+4m$ in the middle, so we'll use $-6m$ and $+10m$

So we have: . $4m^2 - 6m + 10m - 15$

Factor by grouping: . $2m\underbrace{(2m-3)}_\downarrow + 5\underbrace{(2m-3)}_{\hookleftarrow\quad}$

Factor out $(2m-3):\quad(2m-3)(2m+5)$

**Quick** · November 25th 2006, 08:28 PM

Originally Posted by Soroban

Hello, shenton!

Divide 60 by 1,2,3,4,... and keep the ones that "come out even".

$60 \div 1\:=\:60\quad\Rightarrow\quad 1\cdot60$
$60 \div 2 \:=\:30\quad\Rightarrow\quad 2\cdot30$
$60\div 3 \:=\:20\quad\Rightarrow\quad 3\cdot20$
$60 \div 4\:=\:15\quad\Rightarrow\quad 4\cdot15$
$60 \div 5\:=\:12\quad\Rightarrow\quad 5\cdot12$
$60 \div 6 \:=\:10\quad\Rightarrow\quad 6\cdot10\;\;\leftarrow\text{ a difference of 4 !}$

We want $+4m$ in the middle, so we'll use $-6m$ and $+10m$

So we have: . $4m^2 - 6m + 10m - 15$

Factor by grouping: . $2m\underbrace{(2m-3)}_\downarrow + 5\underbrace{(2m-3)}_{\hookleftarrow\quad}$

Factor out $(2m-3):\quad(2m-3)(2m+5)$

I prefer my method, no guess-and-check

**shenton** · November 25th 2006, 08:59 PM

That split middle term method is awesome. I tried it and I do get the answer (your arithmetic is right).

I tried the method on another question and it works!

20a^3 + 54a^2 - 56a
2a(10a^2 + 27a - 28)

Applying the formula, the middle terms are -8 and 35.

2a(10a^2 - 8a + 35a - 28)
2a(10a^2 - 8a) + (35a - 28)
2a(2a(5a-4) + 7(5a-4))
2a (5a-4) (2a+7) -> Factor

That works! Thanks.

**shenton** · November 25th 2006, 09:10 PM

Things are really simpler than it looks.

-6 +10 = 4, I don't know why I never thought of that. I think the coefficient 4 of m^2 freaks me out.

Even if I got that, I don't think I can work out the final answer since I didn't know you can do "grouping" to find the factor. Thanks for teaching that.

I think your method is simpler to use to solve this question.

But when it comes to a question like:
20a^3 + 54a^2 - 56a
2a(10a^2 + 27a - 28)

I think it would be easier to use Quick's method.

Would you agree on that or do you think I should keep dividing as how you showed me?

Thanks.

**shenton** · November 25th 2006, 09:19 PM

Two more tough questions:

16a^2b^2 -25
product = 16 x -25 = -400
sum=no middle term??

Answer key: (4ab +5) (4ab -5)

48x^2 - 27y^2

product = what to multiply??
sum=no middle term??

Answer key: 3(4x + 3y) (4x - 3y)

How do I solve these two questions (what are the steps)?

Thanks as usual.

mphair · November 25th 2006, 11:26 PM

the first problem you gave:
16*a^2*b^2 - 25
is a classic difference of squares problem.
the definition you will want to use is that
a^2-b^2=(a+b)(a-b)
note that this is a standard definition and the a's and b's in the problem are not the same as the a's and b's in the definition.

now take a look at the problem...the first term: 16*a^2*b^2
notice how each of those numbers is a perfect square?
16 = 4*4
a^2 = a*a
b^2 = b*b

now take a look at the second term:
25
notice how this, too is a perfect square?
25 = 5*5

when we have two perfect squares subtracting eachother, we call this a difference of squares and we can apply the definition.

so we factor the problem to two terms: 4ab (which is the square root of 16a^2b^2) and 5 (which is the square root of 25).
plugging them into the definition:
(4ab+5)(4ab-5)

for the second one...notice how there is a three in common with the coefficients of each term. 48 = 3*16 and 27 = 3*9
so with a 3 factored out, the problem becomes very similar to the first problem in which we apply the difference of squares definition and factor accordingly.

hope that helps
-mphair

**topsquark** · November 26th 2006, 04:25 AM

Originally Posted by Quick

I have a method to split the middle!

Given an expression: $ax^2+bx+c$

The middle terms are: $\frac{b\pm\sqrt{b^2-4ac}}{2}$

Note: it looks like the quadratic formula, but it doesn't have an "a" in the denominator and it's just "b" to start, not negative "b"

Substitute: $\frac{4\pm\sqrt{4^2-4(4)(-15)}}{2}$

Thus: $\frac{4}{2}\pm\frac{\sqrt{16+240}}{2}$

Therefore: $2\pm\frac{\sqrt{256}}{2}$

Then: $2\pm\frac{16}{2}$

Thus: $2\pm8$

So you can rewrite $4m^2 + 4m - 15$ into $4m^2-6m+10m-15$

So Factor: $2m(2m-3)+5(2m-3)$

Then add: $(2m+5)(2m-3)$

Quick, your method certainly works, but as I pointed out to you before, I think you are doing too much work in using it.

In this case, we wish to factor $4m^2 + 4m - 15$ .

So solve:
$4m^2 + 4m - 15 = 0$

$m = \frac{ -4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot (-15)}}{2 \cdot 4}$

$m = \frac{-4 \pm 16}{8} = \frac{3}{2}, \, -\frac{5}{2}$

So a quadratic polynomial in m with these roots is:
$\left ( m - \frac{3}{2} \right ) \left (m + \frac{5}{2} \right )$

Multiply each factor by 2 to clear the fractions:

$(2m - 3)(2m + 5)$

(The only thing you need to check is that the $m^2$ coefficient is correct. If not you will need to multiply by another factor to correct it.)

I suppose the main difference is a matter of style, but this method is more direct as it doesn't involve factoring at the end of the problem. It is also (like your method) kind of putting the cart before the horse: both of these methods depend on the quadratic formula which is designed to solve quadratics. The only reason to factor a quadratic is if you either don't know the quadratic formula or are trying to avoid using it. If you are going to solve a quadratic by factoring and don't wish to use the quadratic formula using either of these methods is kind of silly.

-Dan

**Quick** · November 26th 2006, 05:02 AM

Originally Posted by topsquark

Quick, your method certainly works, but as I pointed out to you before, I think you are doing too much work in using it.

In this case, we wish to factor $4m^2 + 4m - 15$ .

So solve:
$4m^2 + 4m - 15 = 0$

$m = \frac{ -4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot (-15)}}{2 \cdot 4}$

$m = \frac{-4 \pm 16}{8} = \frac{3}{2}, \, -\frac{5}{2}$

So a quadratic polynomial in m with these roots is:
$\left ( m - \frac{3}{2} \right ) \left (m + \frac{5}{2} \right )$

Multiply each factor by 2 to clear the fractions:

$(2m - 3)(2m + 5)$

(The only thing you need to check is that the $m^2$ coefficient is correct. If not you will need to multiply by another factor to correct it.)

I suppose the main difference is a matter of style, but this method is more direct as it doesn't involve factoring at the end of the problem. It is also (like your method) kind of putting the cart before the horse: both of these methods depend on the quadratic formula which is designed to solve quadratics. The only reason to factor a quadratic is if you either don't know the quadratic formula or are trying to avoid using it. If you are going to solve a quadratic by factoring and don't wish to use the quadratic formula using either of these methods is kind of silly.

-Dan

Topsquark's Method:

1) $4m^2 + 4m - 15=0$

2) $m=\frac{-4\pm\sqrt{4^2-4(4)(-15)}}{2(4)}$

3) $m = \frac{-4 \pm 16}{8} = \frac{3}{2}, \, -\frac{5}{2}$

4) $\left(m+\frac{3}{2}\right)\left(m-\frac{5}{2}\right)$

5) $(2m+3)(2m-5)$

Quick's method:

1) $4m^2 + 4m - 15$

2) $4m^2 + \left(\frac{4\pm\sqrt{4^2-4(4)(-15)}}{2}\right)m - 15$

3) $4m^2 -6m +10m - 15$

4) $2m(2m-3)+5(2m-3)$

5) $(2m+5)(2m-3)$

Oh yeah, a ton more work than necessary

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