Question 1

How do you simplify an equation

Accepted Answer

You just need to combine like terms. Example- t to the fourth plus t to the fourth; 2w squared plus w squared. I hope this helps and God bless!

Question 2

Am I ever going to need this type of math in the real world?

Accepted Answer

I mean, you'll need it for the SAT or the ACT - the math in those is primarily Algebra 2, and stuff like this, but as far as the real world, it depends on your job... but we have computer software to do this stuff nowdays :)

Question 3

in the vid they never clearly said which goes first they just said its the only one cubed and do I do y squared first or y alone first

Accepted Answer

Generally, you should write your answer in standard form.  This would have the term with y^2 first, then y, then the constant.  However, it is usually not required unless the instructions specify your answer must be in standard form.

Question 4

So, you can apply the distributive property to one of these equations, or you can use the graph/colored boxes thing? Which way is more efficient?

Accepted Answer

There is FOIL, double distribution, and box method as 3 ways of doing the exact same thing, they all give the same 4 terms with the middle terms usually combined.  FOIL is probably less efficient because it is limited to multiplying two binomials.  The box method may require more writing, but might make more sense in exactly what is going on.  Double distribution and box method can easily be expanded to larger polynomials in the future.  Efficient may be in the eye of the beholder, so find which one works best for you.  Many teachers will require you to learn all so that you can make an educated decision which is best, and if you do it enough and have a pretty good math brain, you will be able to do it in your head.

Question 5

Most Confusing thing ever!!

Accepted Answer

Let's say you have an equation like
(x + 2)(3x^3 - 7 + 2x^2)

First start by multiplying x by each of the values in the trinomial.

x * 3x^3 = 3x^4
x * -7 = -7x
x * 2x^2 = 2x^3

So far, you have 3x^4 + 2x^3 - 7x

Next, you multiply 2 by each of the values in the trinomial.

2 * 3x^3 = 6x^3
2 * -7 = -14
2 * 2x^2 = 4x^2

From this you get 6x^3 + 4x^2 - 14

Now, we use the trinomial that we got from multiplying by x and combine like terms.

3x^4 + 2x^3 - 7x
6x^3 + 4x^2 - 14

There is only one term that has an exponent of 4, so we start our equation with that term.

3x^4

We then add 2x^3 and 6x^3 to get 8x^3

We now have 3x^4 + 8x^3

We only have one term with an exponent of 2, so we add that to our expression

3x^4 + 8x^3 + 4x^2

There is only one term with an exponent of 1, so we add (or rather, subtract) that to our equation

3x^4 + 8x^3 + 4x^2 - 7x

Finally, we only have one constant term (term without a variable or exponent), so we add that to our expression, and we get our final expression.

3x^4 + 8x^3 + 4x^2 - 7x - 14

Hopefully that helps!

Question 6

What would be considered the fastest method of multiplying long polynomials?

Accepted Answer

Great question!  Yes there is a shortcut, which efficiently combines all the terms for each power of x.

If only one variable, say x, appears in the two polynomials to be multiplied together, then you can use a technique similar to the Vedic multiplication arithmetic technique called vertical and crosswise (which you can look up online to get the idea).  Write the terms of each polynomial in order of descending powers of x. For any missing exponents on x, use a coefficient of zero.

Example:  multiply (6x^4 + 3x^3 - 5x + 7) by (2x^3 - 4x^2 - x + 8).

Write
(6x^4 + 3x^3 + 0x^2 - 5x + 7)
(0x^4 + 2x^3 - 4x^2 - x + 8).

These two polynomials are now each written with five coefficients.

The idea is to multiply the first coefficient by the first coefficient, then cross multiply the first two coefficients by the first two coefficients, then the first three by the first three, then the first four by the first four, then the first five by the first five, then the last four by the last four, then the last three by the last three, then the last two by the last two, then finally the last one by the last one.

There are no terms of degree 9 or higher.
Coefficient of x^8 is 6(0) = 0.
Coefficient of x^7 is 6(2) + 3(0) = 12.
Coefficient of x^6 is 6(-4) + 3(2) + 0(0) = -18.
Coefficient of x^5 is 6(-1) + 3(-4) + 0(2) + (-5)(0) = -18.
Coefficient of x^4 is 6(8) + 3(-1) + 0(-4) + (-5)(2) + 7(0) = 35.
Coefficient of x^3 is 3(8) + 0(-1) + (-5)(-4) + 7(2) = 58.
Coefficient of x^2 is 0(8) + (-5)(-1) + 7(-4) = -23.
Coefficient of x is -5(8) + 7(-1) = -47.
Constant term (coefficient on x^0) is 7(8) = 56.

The answer is
12x^7 - 18x^6 - 18x^5 + 35x^4 + 58x^3 - 23x^2 - 47x + 56.

Have a blessed, wonderful day!

Question 7

is this apart of the algebra 2 course

Accepted Answer

I would say this is something you learn in Algebra 1. This seems more like a review in Algebra 2.

Question 8

What do you think is the best way to multiply and simplify a problem?

Accepted Answer

In my opinion, the area model is easier because I can see out all the terms and not get things mixed up. I tend to get numbers mixed up and write things wrong when I do the distributive property. However, it is your preference in the end!

Question 9

when you are multiplying without a coefficient how do you deal with the exponent.

Accepted Answer

Please give an example to clarify.  Everything has a coefficient.  For example:  x = 1x.  The coefficient =1.
So, I don't understand what you mean.

Question 10

I got  a negative 6 at the last equation and how is it positive 6

Accepted Answer

It should be positive 6 because when you distribute the - 6 to the - 1, the product is positive 6.

Get ready for Precalculus

Course: Get ready for Precalculus > Unit 2

Multiplying binomials by polynomials review

Example

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