Question

How do you factor 7m^2+6m-1?

Answer 1

Use the quadratic formula\[ax^2+bx+c=0\]\[x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]\[(x-x_1)(x-x_2)=0\]

Answer 2

What is the quadratic formula?

Answer 3

my second line is the quadratic formula

Answer 4

isn't that for when you're trying to find the slope of a line.

Answer 5

The quadratic formula is used for finding the solutions \(x_1, x_2\) to the quadratic polynomial (a parabola),

but you can also use if for factoring

Answer 6

A parabola can be found using trinomial?

Answer 7

yes

Answer 8

how do you do that?

Answer 9

the first step is to realise that \(m\) is like \(x\)

now compare your expression , with my top line,

what are \(a,b,c\)?

Answer 10

Can you like give me an example using numbers?

Answer 11

If you wanted to actually factor without solving first:

\[7m^2+6m-1\]7*-1 = -7, find factors of -7 that sum to 6. -1,7 are said factors.
\[7m^2 +7m -1m - 1\]\[(7m^2+7m)-(m+1)\]\[7m(m+1)-1(m+1)\]\[(m+1)(7m-1)\]

Answer 12

oh I get it now. That's what I did and I got the same answer.

Answer 13

say you had this expression \(y^2-2y-3\)

comparing with the polynomial we see that \(a=1\), \(b=-2\), \(c=-3\)

now plugging these values into the quadratic formula
\[y_{1,2}=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-3)}}{2(1)}\\\qquad=\frac{2\pm\sqrt{4+7}}{2}\\\qquad=\frac{2\pm\sqrt{16}}{2}\\\qquad=\frac{2\pm4}{2}\\\qquad=\frac{2+4}2,\frac{2-4}{2}\\\qquad=\frac{6}{2},\frac{-2}{2}\\\qquad=3,-1\\~\\
y^2-2y-3=(y-3)(y+1)\]

Answer 14

Another way to factor this is to notice that:
\[7m^2 +6m-1 = 7m^2 +(7-1)m-1=(7m^2 +7m)-(m-1) \]
=\[7m(m+1)-(m+1)=(7m-1)(m+1)\]

Answer 15

So a parabola (which this is) will have crossings of the x-axis at the places where the product terms = 0, or \(m = -1, m = 1/7)\). Because the coefficient of the \(m^2\) term is positive, it is a parabola that opens upward. The vertex will be on a line perpendicular to the x-axis, midway between the two crossings of the x-axis, because parabolas are symmetric about the perpendicular line between vertex and directrix.

Answer 16

Oops: The last (m-1) in the first line should be (m+1)

Answer 17

changing (or forgetting to change) the sign in the second group is where I almost always make my mistakes in this, too :-)

Answer 18

I forgot to multiply 7 and 1. And I thought it didn't make sense.

Answer 19

Remembering the quadratic form as @UnkleRhaukus used is a good way to check your work, too. If you do a problem two different ways and get the same answer, the probability that you did it correctly is pretty good.

Answer 20

I have never learned the first part of the quadratic formula.

Answer 21

quadratic form is definitely the one I reach for if it's a real-world problem where the numbers aren't "nice" like they usually are in textbook examples.

Answer 22

Yes the numbers aren't nice even using quadratic formula or the Magic X.

Answer 23

I always use the quadratic formula because it is one of the few formulae i have committed to memory and never have to look up to check, and it always works,

Also i have trouble seeing how to split up that middle term, in the method used my whpalmer4 & RajeshRathod, it just isn't as intuitive (for me personally).
But if you prefer that method and can see the factors as quickly as them, i would recommend you use that method.

I would also note that if you use the quadratic formula method then there is sometimes another step (as in this case)

you would get \(m_{1,2}=\dfrac{-6\pm8}{14}=\dfrac{2}{14},\dfrac{-14}{14}=\frac17,-1\)

so the factor would come out as \[\qquad (m-\tfrac1{7})(m+1)=0\]multiplying by seven, to get rid of the fraction
\[\qquad(7m-1)(m+1)=0\]

Answer 24

ok.

Answer 25

I do them just often enough to remember how, but not often enough to do them as quickly as I can factor things where the coefficient of the x^2 term = 1 :-)

Being able to do it quickly depends on being able to factor the product of the leading and trailing coefficients in a hurry, of course. When the answer is an easily-recognizable prime number like it was for the original problem, that doesn't take long!

Answer 26

ok.

Answer 27

For \(y^2-2y-3\) we also have an easy case: product is -3, we need a sum to -2, so -3 and 1 are the choice. \[y^2-2y-3 = (y+1)(y-3)\]Or if you went to the trouble of splitting:\[y^2+y - 3y -3 =( y^2 + y )- (3y+3) = y(y+1)-3(y+1) = (y+1)(y-3)\]

Answer 28

wow that looks complicated.

Answer 29

it does? I thought that's what you've been doing all night!

Answer 30

no I haven't learned that.

Answer 31

Do a few hundred and it'll get easier :-)

Answer 32

but I don't know how to do that.

Answer 33

but that's what we did in the problem earlier...\(8x^2+32x-12x-48\)
\[8x^2+32x -(12x+48) = 8x(x+4)-12(x+4) = (8x-12)(x+4)\]

Answer 34

oh grouping it?

Answer 35

Yes, that's what I was doing in the second line ("or if you went to the trouble of splitting")

Answer 36

that's what you're suppose to do. you're suppose to split the middle coefficient.

Answer 37

well, if you have to factor something where the coefficient of the \(x^2\) term is 1, you don't have to bother: you just factor the constant term, and pick the pair of factors that sums to the middle term.

For example:
\[x^2-6x+9\]Factors of 9 are 1*9,3*3,-1*-9,-3*-3
-3+-3 = -6, so that's the one we want. Factoring is\[x^2-6x+9 = (x-3)(x-3)\]

Answer 38

Or \[x^2+5x+6\]Factors of 6 are 1*6, 2*3, 2+3 = 5, so 2 and 3 are what we want. \[x^2+5x+6 = (x+2)(x+3)\]

Answer 39

As I was trying to illustrate before that other guy said I was doing it wrong,

\[(ax+b)(x+c) = ax^2 + acx + bx + bc = ax^2 + (ac+b)x + bc\]
We multiply first and last term to get \(abc\), then we look for factors of \(abc\) that give us \[ac+b = \text{<middle term>}\]

For example: \[2x^2+5x+3\]Product of outer terms is 2*3 = 6. We need sum of factors to equal 5, so 2*3. We can't just write \[(x+2)(x+3)\]because the first term won't be \(2x^2\). So we need to write \[(2x+\text{<something>})(x+\text{<something else>})\]Because one of them gets multiplied by 2, we have to divide it by 2 before writing it in the factoring. We can't divide 3 by 2, so we divide 2 by 2 instead. A bit of trial-and-error here.
\[(2x+3)(x+1) = 2x*x + 2x*1 + 3*x + 3*1 = 2x^2 + 5x + 3\]

Answer 40

Pay attention to what happens as you multiply polynomials, and you'll have an easier time reversing the process.

Answer 41

I get it now.

Answer 42

Great!

Answer 43

yay.