Suppose you divide a polynomial by a binomial. How do you know if the binomial is a factor of the polynomial? Create a sample problem that has a binomial which IS a factor of the polynomial being divided, and another problem that has a binomial which is NOT a factor of the polynomial being divided.
I used: (x^2 + 2x + 1)(x - 4) where x - 4 is the factor (right??)
But for part 2, so... can you just do something like: (x^2 + 2x + 1 + 3)(x - 4) to get a different factor? I put in a + 3. Does that change the factor?
Did I do this correctly? Thanks
This is called the factor theorem.
If the division by some binomial leaves a remainder of zero, then that binomial is a factor
the same is true for numbers.
e.g. 12÷3 = 4 with Remainder of 0
so 3 is a factor of 12
12÷5 = 2 with remainder of 2, so 5 is NOT a factor of 12
I have no idea what you are doing in that second part, you will have to divide
(x^2 + 2x + 1) by (x - 4)
Do you know how to do long algebraic division?
Do you know how to do synthetic division?
can u give the proof of factor theorem
To determine if a binomial is a factor of a polynomial, you can use the remainder theorem or perform polynomial long division.
In the first problem, you correctly identified the binomial (x - 4) as a factor of the polynomial (x^2 + 2x + 1)(x - 4). This means that when you divide the polynomial by the binomial, you will get a quotient without a remainder.
Let's perform polynomial long division using the example you provided:
Dividend (polynomial being divided): x^2 + 2x + 1
Divisor (binomial): x - 4
x + 6
___________________
(x - 4) | x^2 + 2x + 1
- (x^2 - 4x)
____________
6x + 1
- (6x - 24)
___________
25
The quotient is x + 6, and the remainder is 25. Since there is a remainder, the binomial (x - 4) is not a factor of the polynomial (x^2 + 2x + 1).
In your second problem, you added a constant term (+3) to the polynomial (x^2 + 2x + 1) before dividing it by the binomial (x - 4). This does change the polynomial, but the addition of a constant term does not change the factor (x - 4) itself.
So, in the second problem, the binomial (x - 4) is still the same factor you are dividing by. However, we need to perform the division to confirm if it is a factor or not.
To determine if a binomial is a factor of a polynomial, you can use the Remainder Theorem or perform polynomial long division. The Remainder Theorem states that if you divide a polynomial P(x) by a binomial (x - a) and the remainder is zero, then (x - a) is a factor of P(x). In polynomial long division, if the division process results in no remainder, then the binomial is a factor.
Let's go through the sample problems you provided:
1. Polynomial: P(x) = (x^2 + 2x + 1)(x - 4)
Binomial: x - 4
To check if (x - 4) is a factor, you can either use the Remainder Theorem or perform polynomial long division. Let's use polynomial long division:
x + 6
----------------
x - 4 | x^2 + 2x + 1
- (x^2 - 4x)
------------
6x + 1
- (6x - 24)
---------
25
The remainder is 25, which is not zero. Therefore, (x - 4) is not a factor of the polynomial P(x).
2. Polynomial: P(x) = (x^2 + 2x + 1 + 3)(x - 4)
Binomial: x - 4
You correctly added a constant term (+3) to the polynomial. Now, let's perform polynomial long division to check if (x - 4) is a factor:
x + 6
----------------
x - 4 | x^2 + 2x + 4x + 4 + 3
- (x^2 - 4x)
--------------
6x + 7
- (6x - 24)
-----------
31
Again, the remainder is 31, which is not zero. Therefore, (x - 4) is not a factor of the polynomial P(x).
In both cases, the remainders were nonzero, indicating that the binomials (x - 4) were not factors of the polynomials being divided.