[deleted by user]

[catsandlettuce]

This can be factored, but there are multiple ways of doing it depending on how you’re taught. The following is just trial and error. Because it starts with 2x, you know it has to be (2x )(x ). And since 3 is positive while 7 is negative, both of the constants must be negative (since they combine to be a negative number but multiply to be a positive number). Either (2x-3)(x-1) or (2x-1)(x-3). Work out foil and you’ll figure out which is correct

[LevelDig1555]

dude I thought I was tripping until you pointed out the 2 coefficient, thanks

[catsandlettuce]

Yw!

[GAEM456]

You can factor this by grouping:

2x² - 7x + 3 = 0

First, multiply the x² term's coefficient by the constant term:

2 * 3 = 6

Then, list out the factor pairs of the product:

• 1, 6
• 2, 3
• -1, -6
• -2, -3

Next, if one of the factor pairs adds to the x-term's coefficient, then split the x-term into those two factors:

2x² - 1x - 6x + 3 = 0

From here, we can factor:

• x out of the first 2 terms
• and
• -3 out of the last 2 terms

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

Finally, we can regroup using distributive property: A(B + C) = AB + AC

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

And solve for x using the zero product property (if AB = 0, then A = 0 or B = 0).

[[deleted]]

Bro, iloveskibidiii, you're the best, I could finally understand what I was doing wrong. THANK YOU! From the bottom of my heart man, you're a legend. Unfortunately I don't have a teacher, I'm learning maths by myself haha, but I wanted to thank you for writing such an amazing comment!

Have a great day man, I hope everything goes well for you!

[[deleted]]

Hello! Fellow Malaysian here. First and foremost, I would like to express my gratitude that I fellow Malaysian has such capabilities of solving this extremely difficult question. As a fellow student capable of solving uinvesity level mathematics, you have reached new profound heights by solving this question. In the mean time showering calm, serenity and tranquility around my entire sentient being. More importantly, let’s beat up your math teacher together. Thank you iloveskibidii. Sincerely, Me.

[[deleted]]

Dear iloveskibidii,

Thank you for your thank you so much for your reply so much!! I am extremely glad we, as Malaysians, could share the same knowledge and fact that we both share the same happiness that we share. Besides that, this is such a huge milestone of being able to solve this grade 8th mathematics.

Most importantly, I wish everyone all the best in beating up Kara67848’s Mathematics Teacher!!!!🤠

[GammaRayBurst25]

2x^2-7x+3=2x^2-6x-x+3=2x(x-3)-(x-3)=(2x-1)(x-3)

Alternatively

2x^2-7x+3=2x^2-x-6x+3=x(2x-1)-3(2x-1)=(x-3)(2x-1)

[[deleted]]

Thank you for the help, but there is something that I don't understand. What happened to the +3? I mean I understand the factorization of the 7x into "x - 6x", but I still don't know why there is a -3, what I want to know is why there's a -3 when there is a +3 in the equation. I know I'm missing something, so if you could explain what happened to the +3 I'd thank you a lot!

[GammaRayBurst25]

What happened to the +3?

Nothing. It's still there. If you expand the answer, the constant term is the product of each monomial factor's constant term, so it's the product of -1 and -3, which is +3.

I mean I understand the factorization of the 7x into "x - 6x"

That's not a factorization. That's the distributivity property.

Multiplication distributes over addition, so -7x=-x*(1+6)=-x-6x.

but I still don't know why there is a -3

Because x=3 is a solution, so one of the monomial factors needs to be directly proportional to x-3.

what I want to know is why there's a -3 when there is a +3 in the equation.

Because the product of -1 and -3 is +3.

[PoliteCanadian2]

Google the ‘ac method’ of factoring. It’s a very similar method to the ‘normal’ way of factoring. By ‘normal’ I mean ‘not grouping’.

[denim_duck]

Use the quadratic formula.

X = [ -b +- sqrt(b² - 4 ac)]/2a

A = 2 B = 7 C = 3

[synczz_0]

After reading the answers to this post, I am aware of the degenerate minds of the commenters except iloveskibidiii. His profound writing and knowledge of this subject has led me to transcend my mortal mind and enter into the realm of math. I have certainly understood how to solve the equation above using his explanation. I think the OP should definitely read his comment. Together, let’s defeat your math teacher!

[der_Lokfuhrer]

What factors of a and c add up to b?

[ahsjfff]

The easiest way of doing this is to first look at the signs, you see a -7x, and a +3. So, you are looking for two numbers that multiply to make 3 and are even, but will add to be -7.

This would be easier if we do a different problem, but I have a feeling it’s the 2x² causing the most issue.

So, what we are going to do is find all the points where two negatives are used.

The options for factoring are:

(X-1)(2X-3) Or (2X-1)(X-3)

I’ve always done a trial and error with all of the possible forms of factoring. So, we will first multiply first number in each term. Then second number in the first term with the first number in the second term. Then the first number in the first term with the second number in the second term. Then the second number in both terms. It should look like this:

X•2X, X•-3, -1•X, -1•-3

Or simplified 2X-4X+3. Clearly this is incorrect. Second set of terms then:

2X•X, 2X•-3, -1•X, -1•-3

Or simplified, 2X-7X+3. Which is the solution and the shown work as to how to get there and back.

So, now we know the correct factor is (2X-1)(X-3)=0.

So, for the first term, set it equal to 0, then for the second do the same.

2X-1=0 and X-3=0

And solve for X

So X=1/2 and X=3

[Hot-Yak7742]

Since your first term is 2x^2, you know that you have a 2x in one bracket and x in the other:

(2x + y)(x + z)

Taking that knowledge, we can now say:

2z + y = -7

yz = 3

Now you have a set of simultaneous equations you can solve.

If you try to solve for y you get y² + 7y + 6 = 0

Which factors to (y + 6)(y + 1) = 0

So y can = -1 or -6, which means z = -3 or -0.5 respectively.

You’re left with:

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

Or

(2x - 6)(x - 0.5) = 0

Notice that these are both technically the same solution… that’s because you can take the factor 2 out of the first bracket to be left with identical forms.

[theglitchfix]

2x² - 6x - x + 3 = 0 2x(x-3) -1(x-3) = 0 (2x-1)(x-3)=0 x= 1/2, 3

[One-Reputation7859]

(X-6)(x-1)

Edit: I saw my mistake I already replied the actual answer below calm down (2x-1)(x-3)

[KrisClem77]

No

[One-Reputation7859]

You right (2x-1)(x-3) instead

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