How Do I Multiply These Polynomial Equations?

[GodBloo]

So basicly our teacher taught us in high school how to find the product of some equations but I do not understand it very well and I need someone to teach me how to solve this basic problem.

The Equation is : (3x^2-4x+1)(4x^2+x-2)

I do not know how to find the product of that problem can anyone please help me with it?

 

[Mentallic]

So basicly our teacher taught us in high school how to find the product of some equations but I do not understand it very well and I need someone to teach me how to solve this basic problem.

The Equation is : (3x^2-4x+1)(4x^2+x-2)

I do not know how to find the product of that problem can anyone please help me with it?

For every term in the first factor, multiply it by the entire second factor. For example,

[tex](a+b)(c+d)=a(c+d) + b(c+d)[/tex]

 

[HallsofIvy]

You do the same thing with polynomials that you do with numbers. If you were asked to multiply 233 by 123, you would first multiply 233 by the first digit in 123, 3
699. Then multiply by 2: 466. Finally multiply by 1: 233

You would position those as
___233
__x123
______
__699
_466
233
______
28659

That spacing is because 123= 100+ 20+ 3 so you are actually multiplying (100+ 20+ 3)233= 100(233)+ 20(233)+ 3(233) (that's the "distributive law").

Similarly to multiply (3x^2-4x+1)(4x^2+x-2) think of it as 3x^2(4x^2+ x- 2)- 4x(4x^2+ x- 2)+ 1(4x^2+ x- 2). You can do each of those by using the "distributive law" again:
3x^2(4x^2+ x- 2)= (3x^2)(4x^2)+ (3x^2)(x)+ (3x^2)(-2)= 12x^4+ 3x^3- 6x^2.
-4x(4x^2+ x- 2)= (-4x)(4x^2)+ (-4x)(x)+ (-4x)(-2)= -16x^3- 4x^2+ 8x
1(4x^2+ x- 2)= (1)(4x^2)+ (1)(x)+ 1(-2)= 4x^2+ x- 2.

Now add those combining "like terms" (terms with the same power of x). The only "x^4" term is 12x^4. The "x^3" terms are 3x^3 and -16x^3- their sum is -13x^3. The "x^2" terms are -6x^2, -4x^2, and 4x^2. They add to -6x^2. The "x" terms are 8x and x. They add to 9x. Finally, the only "constant term" is "-2". That is, the product is 12x^4- 13x^3- 6x^2+ 9x- 2.

In a nutshell, multiply each term in one polynomial by each term in the other, then "combine like terms".

 

[GodBloo]

But what do I do with the exponents? do I add them up?
For exemple (3x^2)(4x^2) = 12x^4
like ^2+^2?

 

[Ssnow]

yes, using power properties in the same bases ## a^{n}\cdot a^{ m}=a^{n+m}##, remember if ##a\not=0 \Rightarrow a^{0}=1## ...

 

[Mark44]

So basicly our teacher taught us in high school how to find the product of some equations but I do not understand it very well and I need someone to teach me how to solve this basic problem.

The Equation is : (3x^2-4x+1)(4x^2+x-2)

I do not know how to find the product of that problem can anyone please help me with it?

Let's get some terminology down. The above is a product, but it is not an equation -- this is an algebraic expression. An equation has = in it, and states that two expressions have the same value. The expressions above are made up of three terms each. The terms are the things being added or subtracted.

To expand (multiply out) the expression above, you need to multiply each term in the second expression by each term in the first expression. All together you will have nine multiplications. Some of these intermediate multiplications will have terms that have the same variable part (such as x³) but different coefficients (the constant that multiplies the variable part). These are called like terms, and can be combined. For example, two of the multiplications are 3x² times x and -4x times 4x². The first product gives 3x³ and the second product gives -16x³. We can combine 3x³ - 16x³ to -13x³, using the distributive property -- am + bm = (a + b)m.

You said that you had studied this in high school. If you are studying this material again, are you working from a textbook? If not, it would be useful to get an algebra textbook, which would list all of the various properties you need to use to carry out the multiplication you're interested in.

 

[GodBloo]

Alright thanks I`ll try buying a good Algebra textbook and hopefully it can help me even more thanks for all the help guys!