Algebra
posted by Kally .
i do not get this
use division to determine whethe the binomial is a factor of f(x)
F(x)=x^3+20x32; x2
yes or no

Algebra 
Reiny
3 is a factor of 12 because 3 divides exactly into 12 leaving no remainder
similarly if x2 is a factor of x^3+20x32 then it must divided exactly into it leaving no remainder.
so...
divide x^3+20x32 by x2 and see what happens.
Respond to this Question
Similar Questions

Algebra 1
Are these right? xy  3x  8y + 24 (3xy)(8y+24) 8y+24 4x^2 + 20x + 5y +xy 36x+(5y+x) 36x5y I don't think so it's right.Your question xy3x8y+24 is itself the answer.It cannot be further calculated bcause it doesn't have the terms 
Algebra Help PLEASE
How is dividing a polynomial by a binomial similar to or different from the long division you learned in elementary school? 
Algebra
•How is dividing a polynomial by a binomial similar to or different from the long division you learned in elementary school? 
AlgebraPlease check
5  sqrt of 20x + 4 >= 3 5 20x + 4 >= 3 54  20x >= 3 1  20x >= 31 20x >= 4 x<= 15 Possible answers are x<=3 1/5 <= x<= 3 x >= 1/5 x >= 0 I think it is 1/5<=x<=3 because both solutions … 
Algebra 2
Use the factor theorem to determine whether the binomial is a factor of the given polynomial. 1. t+sqrt2; P(t)=t^5+t^4+4t+4 2. zi; P(z)=z^7+z^6+z^5+z^4+z^3+z^2+z+1 3. z+2i; P(z)=z^3+z^2+4z+4 Thanks 
algebra 2
this binomial can be factored as the product of a binomial and trinomial. Enter the binomial factor. 64y^15 + 125 
algebra
use polynomial long division to divide 20x^3 10x^2 16x 8 by 5x^2 4 
Algebra Question~!
Suppose you divide a polynomial by a binomial. How do you know if the binomial is a factor of the polynomial? 
algebra1 question
Suppose you divide a polynomial by a binomial. How do you know if the binomial is a factor of the polynomial? 
math
Create a quadratic polynomial function f(x) and a linear binomial in the form (x − a). Part 1. Show all work using long division to divide your polynomial by the binomial. Part 2. Show all work to evaluate f(a) using the function …