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### Course: Digital SAT Math>Unit 4

Lesson 1: Factoring quadratic and polynomial expressions: foundations

# Structure in expressions — Basic example

Watch Sal work through a basic Structure in expressions problem.

## Want to join the conversation?

• Whenever you got algebra question or you're practicing algebra question open desmos graphing calculator in next tab

Now what we can do compare LHS AND RHS
L.H.S = 2x²+16x+30 = 2 (x²+8x+15)
R.H.S = 2(x+b)(x+c)*

Cancel 2 from booth side and we'll get
x²+8x+15 = x² + (b+c) + bc

Now Compare L.H.S and R.H.S
i.e. b+c = 8

This question is quite straightforward what if we got question like what's the value of bc + (b-c) or somthing that we need to find both constant

we can use the help of desmos which is built in graphing calculator inside bluebook app and also we can practice in their webside desmos.com

Now plot this eqn b+c = 8 and bc= 15 as x and y variables
x+y=8 & xy=15
Now,You can see intersecting point
(5,3) ==> (x,y) ==> (a,b)
Now you can get your answer a as 5 and b as 3 .
(16 votes)
• Man guys I just dont get this at all
(9 votes)
• Isn't the answer -8, but not 8 as it mentioned in the video? Because, having formula ax^2+bx+c ( in our example x^2+8x+15), the sum of x1 and x2 must be equal to -b ( in our example b+c=-8) Why then here we have +8 in the example?
(7 votes)
• This can work however -b/a would be equal to -b-c because the roots are -b and -c, not b and c.
-b-c = -8
b+c = 8
(1 vote)
• it just doesn't make complete sense 2 me
(4 votes)
• hi asian here (yes i took that from the nepali person commenting above)

tl;dr: start guessing from the third term.

I know it's hard, but never give up
i'll just do a step-by-step of this question

first of all, ignore the b and c, unless you are already used to factorisation
you'll see, in this expression, all coefficients are odd number. Odd numbers are divisible by 2 so 2(x^2 + 8x + 15)
Now if you remember the equation x^2 + (a+b)x + abx = (x+a)(x+b) which fits the question.
You can start guessing from the second term but that just makes life harder as you ponder if you never should have decided to take SAT in the first place. The better way is to start guessing from the third term. So in this case it's 15.
15 = 3*5
The coefficient of the second term should be 3+5, which in this case, it is.

Here in Japan we learn an efficient method of factorisation called tasukigake so if you want to know what it is tell me.
(3 votes)
• factoring is so hard for me. any suggestions?
(3 votes)
• I know factoring but i sometimes end up with the wrong answer
(3 votes)
• CHECK YOUR WORK!!

don't ask me how i have no clue
(1 vote)
• Y'all... I know how to factor but I feel like these lessons are making me forget TT
(2 votes)
• (p+1)^2 can be factored as (p^2+2p+1^2)?
(2 votes)
• Yes. expanding this equation makes it look like the below equation:
(p + 1) (p + 1)
(p + 1) (p + 1) = p^2 + p + p + 1^2 = p^2 + 2p + 1

Basically, the parentheses mean you are squaring the whole equation.
If you wanted to square P and 1 separately though, it would look more like this:
P^2 + 1^2 (which is simply P^2 + 1).
(0 votes)
• i thought sum of roots of a quadriatic equation was -b/a?? How come we got 8?
(0 votes)
• Here b and c are not the roots of the quadratic equation( roots are -3 and -5), here we are comparing the equation on both sides and writing the values.
(4 votes)
• @ How did he know those two numbers would be b and c before factoring out the whole equation? 😫
(1 vote)
• He knew because he wanted to factorize the equation, and he knew the result would be 2(x+b)(x+c). The question asked him to find b + c, and he had x^2+8x+15. This can also be known as X^2 + (b+c)x+ (bc). We already know that where I put b+c, it is 8, so b+c=8. :)
Hope that helps
(1 vote)

## Video transcript

- [Instructor] We're asked in the equation above, b and c are constants. What is the value of b + c? And they give us the equation over here. So, pause this video and see if you can have a go at that before we work through this together. All right, now let's work through this together. And it looks like what's happening is we have a quadratic on the left and then on the right we have that same quadratic that is factored out, although they don't tell us what b and c are we have to figure that out. So, one way to tackle this is actually let me just rewrite the left-hand side of this. So, it is 2x squared + 16x + 30. And what I wanna do is try to get as close to the form that I have on the right as possible. So, it looks like they factored out a two. So, let me do that. So, this is equal to two times. And if any of this factoring of quadratics is unfamiliar to you, I encourage you to review that on Khan Academy, on The non-ACT portion of Khan Academy to get the basics. But if we factor out of two out of this first term, you're just left with an X squared. You factor out of two out of 16x, you get 8x. And you factor a two out of 30 and you get + 15. And then it looks like what they have done is they have factored this part into x + b X x + c. And the simplest way to factor things is to say, all right, are there two numbers that when I add them, I get eight, and that when I multiply them, I get 15? And those two numbers are actually going to be b and c this is one of our main factoring techniques. So, b + c needs to be equal to eight, and b X c needs to be equal to 15. And if we figure that out, then we can factor completely. Well, we've just actually answered their question, b + c needs to be equal to eight. And so, eight is the answer. Now, let me just factor this out completely, so that you can see that a little bit more completely. I'm using the word completely a lot. So, if I were to factor this out, this is the same thing as two times the two numbers that add up to eight. And when I multiply them, I get 15, let's see three and five seemed to work. So, it's gonna be two times X plus three times X plus five. You can verify that three times five you're gonna get that 15 there. And then when you multiply these two binomials, you're gonna get 3x + 5x, which is going to be 8x. And so, you can see that you can either treat b as three and c is five or b is five and c is three, but either way, b + c is going to be equal to eight.