Solving Quadratics by Factoring

Alright, now that we know what polynomials are, it's time to learn how to solve them! But how can we do that? There's terms of different degrees, it's a mess! Well as it turns out, there are a few techniques we can learn, and let's start with factoring.

Check out "Is This Wi-Fi Organic?", my book on disarming pseudoscience!

[NickHope]

COMPREHENSION #4 SOLUTION
Background: If the first term in the trinomial has a coefficient (i.e. a multiplying number), you need to factorize that coefficient and then look for solutions in all the resulting options. In the case of 6x^2 - 13x - 5, you have to factorize the 6, which can be 6 x 1 or 3 x 2 (you only need to determine the positive factors). Because 6 has 2 possible pairs of (positive) factors, you then have to try solutions for both (6x + m)(x + n) and (3x + m)(2x + n) until you find values for m & n that work. It's basically more work than solving simpler quadratics and there's no shortcut.

Problem: 6x^2 - 13x - 5
1. Factorize -5 to get possible pairs of value for m and n: Either -5 * 1 or 5 * -1
2. Factorize 6 (positives only) to get possible pairs of coefficients for x in the binomials: Either 6 * 1 or 3 * 2.
3. Express the 6 * 1 option as 2 binomials: (6x + m)(x + n)
4. Do just the 'OI' in 'FOIL' (i.e. Outer then Inner) to get the equation to solve for m & n: 6n + m = -13. This saves us having to write out the whole equation each time.
5. Try substituting -5 for m and 1 for n: 6 * 1 + (-5) = 6 - 5 = 1. Nah, not -13.
6. Try substituting -1 for m and 5 for n: 6 * 5 + (-1) = 30 - 1 = 29. Nah, not -13.
7. Try substituting 5 for m and -1 for n: 6 * (-1) + 5 = -6 + 5 = -1. Nah, not -13.
8. Try substituting 1 for m and -5 for n: 6 * (-5) + 1 = -30 + 1 = -29. Nah, not -13.
9. Express the 3 * 2 option as 2 binomials: (3x + m)(2x + n)
10. Do just the 'OI' in 'FOIL' (i.e. Outer then Inner) to get the equation to solve for m & n: 3n + 2m = -13
11. Try substituting -5 for m and 1 for n: 3 * 1 + 2 * (-5) = 3 - 10 = -7. Nah, not -13.
12. Try substituting -1 for m and 5 for n: 3 * 5 + 2 * (-1) = 15 - 2 = 13. Nah, not -13
13. Try substituting 5 for m and -1 for n: 3 * (-1) + 2 * 5 = -3 + 10 = 7. Nah, not -13.
14. Try substituting 1 for m and -5 for n: 3 * (-5) + 2 * 1 = -15 + 2 = -13. YES! 8th time lucky.
15. So 6x^2 - 13x - 5 factors to (3x + 1)(2x + (-5)) which is (3x + 1)(2x -5)
16. Check by FOIL (First Outer Inner Last): 3x * 2x + 3x * (-5) + 1 * 2x + 1 * (-5) = 6x^2 + (-15x) + 2x + (-5) = 6x^2 - 13x - 5

[theguythatdoesmath]

it takes around 5.5 hours to watch the entire algebra series on professor dave explains, but the average high school algebra class (the full year) takes 294 hours

[thegoodlydragon]

Wait, I realized why I had so much trouble with the last practice problem. It's because it wasn't explained that a term like 6x^2 could be split into 2x and 3x. I thought it had to be split into 6x and x because that's how the problem in the tutorial with a coefficient greater than 1 was split up.

[anonymous.youtuber]

Very nice how you explain everything fully. I remember some of my math teachers using tricks without explaining them, in the way a magician never reveals his secrets. With you, math is on stage, not your persona. That ‘s really great. 👍

[andrewkorsten]

This dude is better than each of one's teachers - this is phenomenal! I think that it makes sense to just keep playing these vids in schools and fire most of teachers!

[FrancescoDiSiena]

crazy how the full playlist is 5 hours and just when i arrived in this video i learned more in a few days against 20 years of nothing, and i understood even why it works like this, when instead the little things i knew before i had no clue why it worked like that

[BotOpenAI]

For the last question in Comprehension:
You don't need to split it up. You can still keep it 6x and x, your answer will just be (6x + 2)(x - 2.5)
You have to notice that 2 x -2.5 is the other pair that equals -5.

[slickyy]

Is this correct?

6x²-13x-5

Factorised the middle term(-13x)
Then it became:
6x²-3x-10x-5

Then took common factors(Reverse):

(2x-1)(3x+5)

Is that correct?

[robinwei]

There's an error around 0:2:48. You won't get the pairs (10, -1) or (-10, 1) here. The two factors have to be of the same sign.

[thegoodlydragon]

The last two example problems were really hard. Could you please make a video clarifying how it works when the first term in the trinomial has a coefficient greater than 1?

[xx_xxxxx_xx]

im so immensely impressed by how clear your explanations are. i dont know if i'm going to watch the whole playlist but pls add to the series to make this gold accessible.

[focusiam]

Thank you so much Dave for this amazing list. I never comprehended it at school, only god knows how I passed. I will go all the way till the end and donate because what you do doesn't have a price. Thanks again!

[thedominion]

Is ( -2x + 5 ) ( -3x + 1) also valid for problem four?

[waisalmakaleh]

This helped clear some things for me, my calculus exam is tomorrow. My professor did say you truly start to learn algebra in calculus

[Falco]

considering the quadratic #3, if we are asked to solve for the quadratic =0, couldn't we simplify the terms to x^2 - 2x + 1 and get a factorization of (x-1)^2.

[soakedso]

that dopamine hit when you figure out how to do the last problem

[MiltosPol-qnzh]

When are you planning to start geometry???

[arrowhead]

Alright bros my brain has finally stopped functioning

[toothpaste_tm]

Watching these vids helps me in engineering school xD

[andrethornejr]

I swear to God I love this guy he did better than my teacher would have xd love your vids hopefully I pass my exam tomorrow(Monday 23rd May 2022)