SOLUTION: completely factor the expression a^2+4b-ab-4a
a. (a+b)(a-4)
b. (a-b)(a-4)
c. (a+b)(a+4)
d. prime
If you could please help me understand this ty
Algebra.Com
Question 631724: completely factor the expression a^2+4b-ab-4a
a. (a+b)(a-4)
b. (a-b)(a-4)
c. (a+b)(a+4)
d. prime
If you could please help me understand this ty
Answer by lenny460(1073) (Show Source): You can put this solution on YOUR website!
completely factor the expression a^2+4b-ab-4a
a. (a+b)(a-4)
b. (a-b)(a-4)
c. (a+b)(a+4)
d. prime
The above expression cannot be factored. It is prime.
The answer is d).
obuong3@aol.com
RELATED QUESTIONS
completely factor the expression 2a^3-128
a. 2(a-4)(a^2+4a+16)
b. 2(a^3-64)
c. prime... (answered by jim_thompson5910)
completely factor the expression 4q^2+27r^4
a. 31q^2+27r^4
b. 27r^4(4q^2+1)
c.... (answered by jim_thompson5910)
completely factor the expression 16x^4-81y^4
a. (2x+3y)(2x-3y)(4x^2+9y^2)
b. prime
c.... (answered by Tatiana_Stebko)
completely factor the expression:... (answered by stanbon)
solve the equation 1/4a^2 + 3/4a = 1
a. a =4 and a=1
b. a= -4 and a = 1
c. no solution
(answered by htmentor)
simplify the expression 2ab^4-3a^2b^2-ab^4+a^2b^2
a. ab^4-2a^2b^2
b. 0
c.... (answered by ewatrrr)
a-b/4b-4a (answered by stanbon)
I need help in figuring out the answer to these two problems. I am having a tough time... (answered by user_dude2008)
The average of a,b,c, and d is
(1)a+b+c+d over4
(2)a+b+c+d/4
(3)4a+4b+4c+4d/4... (answered by jim_thompson5910)