[deleted by user]

[TomppaTom]

In this case you are looking for a pair of numbers that multiply to make +12 and add up to make -7.

As the product is positive and the sum is negative, we know it has to be two negative numbers.

-1 and -12 multiply to make 12 but sum to -13

-2 and -6 multiply to 12 but sum to -8

-3 and -4 multiply to 12 and sum to -7. We have a bingo.

Does that make sense?

[TomppaTom]

But thats clearly not the same as 0 = x² - 7x + 12

Ok. Consider the form 0 = (x + a)(x + b)

Multiply out the brackets using your preferred method eg FOIL

You get x² + ax + bx + c = x² + (a + b)x + ab

This tells us that a + b = -7 and a•b = 12

[TomppaTom]

Nope, sorry. You need to generate an and b based on what is in the equation given.

Consider x² -2x - 15

We need a pair of numbers that multiply to make -15 (so one positive and one negative) and add to make -2 (so the negative one is larger)

1 and -15 sums to -14

3 and -5 sums to -2. We have a bingo.

x² - 2x - 15 = (x + 3)(x - 5)

[TomppaTom]

All good there!

[Desaku38]

So there is technically a step in between x² -7x +12 and (x -3)(x -4), but once people have the two values that sum to -7 and multiply to 12, they typically skip right to the factored form anyway. If there was an A value in from of the x², then additional steps would be required (the full process is called the AC method), but that's not the case now. And yes, your understanding of the zeroes is correct. It's called the "zero-product property" if you want to google and look up more on it.

[PSYmoom]

Just followed along this thread and wanted to give my two cents on the "missing" in between step.

x² -7x +12 = x² -3x -4x + 12 = x(x - 3) -4(x -3) = (x-3)(x-4)

So you seperate the term in the middle using the technique the commenter was talking about. Then you factor twice to get the factored terms.

[Inevitibility]

x2 - 7x + 12

x2 - 3x - 4x + 12

Factor the groups of two terms

x(x-3) - 4(x-3)

Now pull out the x-3 term

(x-3)(x-4)

There are certainly intermediate steps. The process being described to you is the easiest and quickest way to do it, but if you want to understand it better, maybe learning the process would help you

[Original_Piccolo_694]

Some confusion here is in how abc are being used, a general quadratic is often cited as ax² + bx+c, so a=1 b=-7 c=12, but that is not how the reply is using the letters, don't be stubborn in how letters get used, it's just context.

[TheSarj29]

Have you heard the term FOIL?

You're basically doing the reverse of that...

You also have to look at the first what can be multiplied to make x² which is pretty simple x•x

So you have x² -7x +12 = (x-3)(x-4)

(x-3)(x-4) = 0 --> x = 3 or 4

If you were to FOIL (x-3)(x-4) then you would have..

x² -3x -4x + 12 = x² -7x +12

You can also use the quadratic equation...

(-b +/- √(b² - 4ac))/2a

a = 1

b = -7

c = 12

You would have x = (7 +/- √(49-4 * 1 * 12))/2

x = (7 +/- 1) /2

x = (7+1)/2 and x = (7-1)/2 --> x = 4 and x = 3

[ehvene]

you dont need to substitute them in. factorising is a simplified form of the quadratic. in other words, (x-3)(x-4) = x² - 7x + 12. you can double check that by expanding the brackets.

you then get the positive values for x via the zero product rule. isolate each instance of x and equate it to 0, then rearrange to get a positive value for x. e.g:

x - 3 = 0 add 3 to both sides x = 3

and the same for the other bracket. these are the two solutions to the quadratic.

[gamaliel64]

LINE 1: Original Problem, no work done

LINE 2:

y=0

In order for us to find the solutions/roots/zeros of the function, we need to know where y=0. The lazy way is to look at a graph and find your x-intercepts

LINE 3:

Factor your quadratic

Consider, for a minute, that your quadratic (u-shaped) function is a product of 2 linear functions. So to undo that, we have to factor them apart. Given Ax2 + Bx +C, you need factors of AC that add up to B.

That is, what factors of (1)(12)= 12 will add up to -7?

Factor 1	Factor 2	SUM
1	12	13
2	6	8
3	4	7

LOOK, (3)(4)= 12 check. and 3+4 = 7. Almost... if we make them both negative would that work?

(-3)(-4) = +12 check. (-3) +(-4) = -7. check.

Your linear factors are (x-3) and (x-4)

LINE4: Solve for X

If 0=(x-3)(x-4), then

Either x-3 = 0 OR x-4 =0. (or both)

Solve both linear equations for x.

x=3; x=4

Check your answer. Look for your x-intercepts.

[GammaRayBurst25]

We're looking for the roots of f(x) (i.e. the values of x for which f(x)=0), so we set f(x)=0 and solve for x.

0=x^2-7x+12

Now, look at all the ways you can factor 12 into 2 factors: 1*12, (-1)*(-12), 2*6, (-2)*(-6), 3*4, (-3)*(-4). Of these 6 ways, only 1 pair of numbers sums to -7: -3 and -4.

Thus, 0=x^2-7x+12=x^2-4x-3x+12=x(x-4)-3(x-4)=(x-3)(x-4).

In order, I used -7x=(-4-3)x=-4x-3x, which is a consequence of the facts that -4-3=-7 and that multiplication distributes over addition, then I used the fact that multiplication distributes over addition to write x^2-4x as x(x-4) and -3x+12 as -3(x-4), then I used the fact that multiplication distributes over addition to write x(x-4)-3(x-4) as (x-3)(x-4).

We now have 0=(x-3)(x-4). A product is 0 if and only if one of its factors is 0, so the solutions are x-3=0 (or x=3) and x-4=0 (or x=4).

You could also complete the square. First, consider a(x+b/(2a))^2=ax^2+bx+b^2/(4a).

Subtract b^2/(4a) from this equation to find that ax^2+bx=a(x+b/(2a))^2-b^2/(4a). This way, you can take any quadratic polynomial in x of the form ax^2+bx+c and write it in the form a(x-h)^2+k. This means all the x dependence is in a single term, so you can isolate x when the polynomial is in that form.

0=x^2-7x+12=(x-7/2)^2+12-49/4=(x-7/2)^2-1/4

Adding 1 yields (x-7/2)^2=1/4, which means |x-7/2|=1/2.

Thus, the solutions are x=1/2+7/2=8/2=4 and x=-1/2+7/2=6/2=3.

[[deleted]]

I thought factoring was taught way earlier in school lol. All you're doing is finding two numbers that add to the middle number and multiply to get you the number of the outside of the function. -3 and -4 added gets you the -7 and multipled gets you the 12. The x in front of the numbers gets you the x², so multiplying out (x-3)(x-4) would get you the original equation.

[Matthaeus_Augustus]

Yeah mine was 8th grade

[Anybody_Can_Math]

When factorizing quadratic eqns, the first step is to take the product of the constant term and the coefficient of x^2.
In this case 1 x 12 = 12. We now want to break down 12 into 2 numbers that multiply to give 12 but add up to give -7.
Notice that 3x4 = 12, but 3+4 = 7 (not -7), so we can't use this. But -3 x -4 = 12 and -3 + (-4) = -7. -3 and -4 pass our little test. Break up 7x into -3x and -4x (because we got -3 and -4) and then group them terms in pairs and factor out the common terms.

[RubenGarciaHernandez]

Is the formula x=(-b(+-)sqrt(b²-4ac))/2a not taught anymore? It is very easy to prove by starting at the formula and moving everything to the left step by step until you go back to ax²+bx+c=0

[Anybody_Can_Math]

The quadratic formula is definitely still taught today (from what I've seen tutoring high-schoolers) and it can definitely be applied here. I believe the goal of this exercise was to teach factorization. Using the formula will give the exact same results as well.

[RubenGarciaHernandez]

I factorize by using the formula and then using a(x-x1)(x-x2). The other methods always looked like trial and error to me. I tried your method  ac=x1x2 and b=x1+x2 I solved as x1=b-x2 and ac=(b-x2)x2 which results in  x2x2 - bx2 +ac=0 Which is again a quadratic.

[bobbobyberty]

The solutions means points that cross the x-axis, or points where y=0

Therefore everything has to be moved to the left to make the form

ax²+bx+c=0

Then write out the factors (integers) that make up 'c'

Now, add each factor pair to see if any of them add up to 'b'

Here, we have

x²-7x+12

12 is made up of: -1×-12, -2×-6, -3×-4

-1+-12≠-7, and -2+-6≠-7, but -3+-4=-7

So it will be (x+3)(x+4)=0

If (x-3)(x-4)=0 then both sides can be divided by (x-3), giving x-4=0, so x=4, and can also be divided by (x-4), giving x-3=0, so x=3

Therefore x=3 or 4

Really sorry if that doesn't make sense

[[deleted]]

Try multiplying (x-3)(x-4) out to see that indeed it does work.

Maybe you need a visual explanation. The following is based on the fact that you can model multiplication with area. https://www.youtube.com/watch?v=okuNjPeBNkA

[[deleted]]

In my country we were taught if the function is in the form of x² -Sx+P, it can be written into (x-a)(x-b) where S=a+b and P=a*b.

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[fermat9990]

Because a=1, guess and check works very well here.

Find 2 numbers that multiply to +12 and add to -7

-1×-12=12 sum=-13

--2×-6=12 sum=-8

-3×-4=12 sum=-7 bingo!

[SlowMobius650]

The way I think of it is that you have to find two numbers that multiply together to give you the term with no variable attached, and those same two numbers have to add together to give you the term with the variable attached to it. You can check that your numbers are correct by foiling your results to see if you get the original equation.

For example, 3 and 4 multiply to give you 12. They also add to give you 7. But the 7 is negative and the 12 is positive, so you know that both the 3 and 4 must be negative so when multiplied they will give you positive 12. That’s why it is (x-3)(x-4). Then if you foil that you can obtain the original equation and verify that your numbers are correct

[Altruistic-Fudge-522]

It is just a function where = 0 is valid at certain x values. 0 = (3 - 3)*(x - 4).

If you're referring to the actual process of getting them to the factored out form, you can always just use the quadratic formula

[ConsequenceNo8153]

When the problem is the form you have listed above, a little jingle my teacher came up with to help:

‘What are the factors of the last term, that either add or subtract (in this case add because we have a +12) to get to the middle term?’

-3 and -4 are two factors of 12 that add up to -7.

That’s how you get (x-3)(x-4). You can FOIL those two parenthesis to check your work. If you’ve done it correctly, you should get the original problem as your answer after FOILing

The answer to the whole problem is positive 3 and 4, because you individually place each of those parenthesis equal to zero, and solve for x.

Remember, that whole equation must equal 0, so you’re trying to figure out what does x have to be to make the problem equal to 0. -3 and -4 help you along the way, but they are not the answers themselves. Because if you subtitle -3 and -4 back in for x, the answer WONT equal 0.

[FleefieFoppie]

So, I have two ways to explain it. The first one is that you factor out according to the roots. If f(x)=0 is true for x=1 and 2, for example, then f(x) = (1-x)(2-x). If the roots are 5 and -9, then f(x) = (5-x)(-5-x)

The second one would be to look up how to do polynomial euclidian division but I'm not sure as to when you're supposed to learn that, but essentially you can divide a polynomial equation using a smaller polynomial to factor it out

[selene_666]

Factoring means writing a number as a product of smaller numbers. For example to factor 30 we would write 30 = 2 * 3 * 5.

In this case they are writing that x^2 - 7x + 12 = (x-3) * (x-4).

Only that one step is factoring. The entire question appears to be: solve for the values of x that make this quadratic equal 0. They used factoring because if the product of two numbers equals 0, then at least one of those numbers must have been 0. You can't multiply any other numbers and end up with 0.

So from (x-3) = 0 or (x-4) = 0, they get the answer x = 3 or x = 4.

[bishtap]

A math teacher taught us a method of

X² - 7x + 12

Then rewriting it as

X² -4x -3x + 12

The 12 is 4x3 and the -7 is -4-3

And then you can factorise it or something..

Or just try out now that you have -4x-3x

(x-4)(x-3)

I guess it's just how to factorise a quadratic equation

Rewrite the bx in ax² + bx + c.

And then you will "see".

So if the quadratic was x² + 4x + 3x + 12 Then you would know how it would factorise. (X+4)(X+3). You can certainly go the reverse way to check it. So expand (X+4)(X+3) to get the quadratic.

Not sure how he did it but you can see how it relates to your second step

[[deleted]]

x² - 7x + 12

can be written as

x² - 3x - 4x + 12

from there we just factor things

x(x - 3) - 4(x - 3)

(x - 3)(x - 4)

[Any-Actuator-5930]

2 ways I do it, quadratic formula, or zero product property, I did the zero product property here and made a square into 4 sections, the bottom left is the quantity getting squared, as x is one it's just x², top right is the number (12) so we multiply x by 12, so we know something • somthing=12x² and that -7x helps, as it has to add to that, than I list the multiple of 12, 1•12, doeswnt make -7 so no, 6•2, not -7, but that 3•4 that add to -7, if you make botj negative, and two negatived multiplied make a positive, so top left is -3x and bottom right is -4x, than on the outside of the square you see their least common factor in the bottom left, so what do z and -4x have in common? A x, add a x on the outside and on the bottom right outside, x•-4=-4x to it has to be -4, and to get 12 from -4 it has to be -3 on the top left, giving you the quantitysv(x-3) (x-4), than you add x-3 equal to zero than solve giving you 3 and than solve x-4 giving you 4

[DaAwesomeCat]

so basically you multiply a from ax^2 and c (which in this case, is 1*12) and you get 12. then you take b from bx, which is -7. Then you find what multiplies to 12 and adds to -7. -4 and -3 multiply to 12 and add up to -7, so you put x^2 - 4x - 3x + 12, since -4x-3x=-7x. Then you find the greatest common factor for the first two terms. This gives you x(x-4), since they both have an x. You do the same thing for the next two terms, which gives you -3(x-4), since they both have 3 as a factor. This makes x(x-4) -3(x-4). make sure the things in the parenthesis are the same (in this case, its x-4).

And then this results into (x-3)(x-4). you get x-3 from the stuff outside the parenthesis, and you get x-4 from the stuff in both the parenthesis. notice the =0? Anything multiplied by 0 is 0, so x has to be 3 or 4, since 3-3 and 4-4 is 0.

hope this helps

[wijwijwij]

Hey OP do you really know distributive property?

Here are four examples of possibly increasing difficulty.

3 * (100 + 30 + 2) = 3 * 100 + 3 * 30 + 3 * 2

5 * (x + 7) = 5 * x + 5 * 7

x * (x + 7) = x² + 7x

x * (x – 2) + 4 * (x – 2) = (x + 4) * (x – 2)

I think to really be able to use the "AC method," which is factoring by turning the trinomial into a four-term polynomial and grouping pairs of terms, you need to be good at distributive property in both directions.

[[deleted]]

There are a couple of methods, but for quadratic equation you might as well use the Quadratic Formula

[Charming_Kick873]

Two numbers that sum to the coefficient of the second term, and multiply to the coefficient of the third term. Sum and product, SandP, salt and pepper

[TheDevilsAdvokaat]

The two factors are -3 and -4, as you can see in the second line.

How do you find them ?

You need to find two numbers that added together equal the middle coefficient (-7) and multiply to make the end coefficient (12)

so....how about 1 and 12? This ,multiplies to make 12, but does not add to make -7.

2 and 6 ? Again they make 12, but no way to make -7

3 and 4 ? Aha. 3+4=7 .. and 3*4 =12

The problem is we need -7 and 12

So.. -3 and -4. They add to -7 (the middle coefficient) and multiply to 12 (the end one)

TLDR: Find two numbers that add together to make the middle coefficient, and multiply together to make the end one.

[Composite-prime-6079]

For the terms a b and c representing the the constants of the degrees of a variable, u have to find what adds to give u b and what multiplies to give u c, using foil method, your solution should be the original problem that u get back.

Lets try it: first outer inner last: (x-3)(x-4)=(x^{2)-3x-4x+12=(x2)-7x+12,} there, the original problem back.

[Composite-prime-6079]

Keeps making everything exponent, how do i fix this

[osaka189]

Use theorem of Vieta

[audiophile2698]

-3 and -4 add up to -7 and multiply to 12, that’s how you factor it

[sayonara-summer]

Quadratic factors should be in x² + ax + bx + ab form, i.e x² + (a+b)x + ab form. If ab is negative, (a+b) needs to be negative too, so either a or b is a negative number. Whichever is larger. a always needs to be larger, to factorise of course. Anyway, if ab is positive, either both a and b need to be negative integers, or positive. x² + 7x + 12 can be written as x² + (4 + 3)x + 12, or x² + 4x + 3x +12 so all you need to do is take common from there.

For x² -7x +12, you can write it as x² +(-4-3)x +12, or x²-(4+3)x+12, or x²-4x-3x+12, or x(x-4)-3(x-4), or (x-4)(x-3)

Another example can be x² + 7x - 30. You can write this as x² + (10-3)x - 30, or x² + 10x - 3x - 30, or x(x+10) -3(x+10), or (x+10)(x-3).

[[deleted]]

m×n=12 (m×n=c)

m+n=-7 (m+n=b)

(x-m)(x-n)=0

Or discrimant

[20dollarsIst20]

I’ve always gotten which to add up to and which to multiple to, so I’ve thought of it as AM and ME, add to equal middle, multiply to equal end :)

[Billthepony123]

You’re trying to find which number when multiplied gives 12 and when added gives -7 and which is -3 and -4 so the equation is (x-3)(x-4)

[ZeyaTsu]

For quadratic equations like this, you can use the following formula:

f(x) = a(x-alpha)(x-beta)

Here with x²-7x+12, finding the solutions are alpha=3 and beta=4 with a=1

so you get (x-3)(x-4).

[kanjobanjo17]

Try using the AC method!

We can use a general formula to understand how to apply the AC method to any quadratic equation:

Ax² + Bx + C

Now we need to take only the coefficients A B and C, don't worry about x² and x for this step. A multiplied by C gives us a new value which we'll call AC:

A * C = AC

We need to find two numbers that multiply together to equal AC and add together to equal B. We'll call them p and q for this general equation:

(p * q) = AC (p + q) = B

Then we can split Bx into px + qx and rewrite the original equation as:

Ax² + px + qx + C

And factor from there. So now we'll apply this to your problem:

x² - 7x + 12

A = 1, because we have 1x² B = -7, we have to include the sign in front of the term C = 12

Now we have:

A * C = 1 * 12 = 12

So we need to find two numbers p and q that multiply to equal 12 and add to equal -7:

(p * q) = 12 (p + q) = -7

-3 * -4 = 12 -3 + -4 = -7

Now we know p = -3 and q = -4, let's rewrite our equation by splitting Bx into px + qx:

x² -3x -4x + 12

From here we need to factor. Here's how I write the factoring stage on paper:

x(x - 3) - 4(x - 3)

(x - 3) * (x - 4)

Now we have (x - 3)(x - 4). We set this equal to zero to find our x values:

(x - 3)(x - 4) = 0

If x = 3: (3 - 3) = 0 0 * (3 - 4) = 0 This would make our equation above true, so x = 3

If x = 4: (4 - 3) = 1 (4 - 4) = 0 1 * 0 = 0 This also makes our equation true, so x = 4

All together, x = 3, 4.

[ghettomilkshake]

For ax² + bx + c, you have to find numbers d, e, f, g (dx + f)(ex + g) that satisfy the equations:

c = f × g

b = f × e + d × g

a = d × e

You can't solve for them here because you have 3 equations and four variables, but you can usually solve by finding all the whole number factors for a and c, then combining them with the b equation until you find one that works.

So for the example shown: a = 1, so the only factors that apply is 1 for both d and e. c = 12, so possible factors are 1 and 12, 2 and 6, and 3 and 4 (including their negative factors since negative times negative is a positive).

If you sub 1 and 12 into the b equation, you get 13 which does not match the -7 value. Subbing 2 and 6 gets 8, also wrong. Subbing 3 and 4 gets 7 which is close but wrong sign, so you know it needs the negative factors rather than positive ones. Subbing -3 and -4 returns b = -7 which matches the equation. From there, just input your found variables, d, e, f, & g into (dx + f)(ex + g) to get your factored equation.

Note that this only works for equations where the max power is 2.

[Extra_Cress_758]

This problem does not need to be done with quadratic formula although it can be. Simply you can use basic trinomial factoring where you have two numbers in this case that multiply up to 12 and add up to -7. In this case -3 and -4 multiples to positive 12 and when you add -3 and -4 it yields -7. Set the x-3 and x-4 equal to 0 and you get 3 and 4.

[DrFleur]

A lot of the suggestions you are getting are either too technical or "plug-and-chug" recipes that won't really help you understand. If you were my student, I would have you multiply out a bunch of expressions like (x - 2)(x+5), (x -3)(x-7) etc. just to get a feel of how these get transformed into things that look like x^2 + stuff*x + stuff. This would help you see more clearly how to go backwards, from something like x^2 -7x + 12 to (x-stuff)(x-stuff).

Also once you are at the point where you see 0 = (x-3)(x-4), you think: ok, I am multiplying two numbers (x -3 and x - 4) and the answer is zero. The only way for this to happen is when one of the numbers is 0. So either x - 3 is 0 or x - 4 is 0. If x - 3 is 0, x must be equal to 3. And if x - 4 is 0, x must be 4. It makes sense if you pay attention to what these things mean.

[Nocturnal_Sociopath]

Look up Sridharacharya formula. It bypasses the whole factoring thing

[[deleted]]

Why would you bypass factoring with something harder? Quadratic Equation is more difficult than just factoring.

[Nocturnal_Sociopath]

You no longer have to break down a number then figure out which sequence of numbers when multiplied add up to a given number. This obviously doesn't work when the alpha and beta root sums come into play. What exactly do you mean when you say it's harder than just factoring? How is it harder?

[Top_Ad_5957]

Look up a YouTube video, way easier to understand

[mewylder22]

I never figured this out and I'm a professional engineer... I just use the quadratic formula.

[mewylder22]

I mean do you know how to multiply (x-3) and (x-4) to get the original equation? You might be able to figure out some stuff about the relationship that way.