|
- How do you factor completely 4x^3-16x^2-9x+36? - Socratic
4x^3-16x^2-9x+36 = (2x-3)(2x+3)(x-4) The difference of squares identity can be written: a^2-b^2 = (a-b)(a+b) We will use this with a=2x and b=3, but first note that the ratio of the first and second terms of the given cubic is the same as that between the third and fourth terms
- Whats the greatest common factor for 27 and 36? - Socratic
The greatest common factor of 27 and 36 is 9 One way to find the greatest common factor (GCF) is to list all of the factors for the numbers Factors are all of the
- How do you figure out the factors of 36? - Answers
Since 36 is a factor of 72, the factors of 36 are included in the factors of 72 What is unusual about the factors of 36 and 12? Since 12 is a factor of 36, the factors of 12 are contained in the
- How do you solve w( w - 5) = 36? - Socratic
Fishing for factors We want to find two numbers which differ by #5# whose product is #36# The numbers #9# and #4# work, in that #9*4=36# and #9-4 = 5# Hence #w = 9# is a solution This is a quadratic equation, so it should have another solution Note that #w = -4# works too, since #(-4)*(-9) = 36# too #color(white)()# Completing the square
- What is prime factorization in expanded form for 36? - Answers
It is: 2*2*3*3 = 36 Q: What is prime factorization in expanded form for 36? Write your answer
- How do you find the solution to the quadratic equation #36=5x+x^2#?
x= -9,4 Factorise the quadratic x^2+5x-36=0 by splitting the middle term x^2 +9x-4x-36=0 The factors would be (x+9)(x-4)=0
- What are the factors for a^6 + 7a^3 - Socratic
#color(white)"ssssssssss"# #= (a+root(3)6)(a^2-root(3)6 x +root(3)36)# #a^6+7a^3+6 = (a+1)(a^2-a+1) (a+root(3)6)(a^2-root(3)6 x +root(3)36)# The two quadratics have no factors over the real numbers Finally, we may be factoring over the complex numbers, as some parts of advanced algebra or precalculus classes sometimes do we continue (Often
- How do you factor 36x ^ { 2} + 113x + 72? - Socratic
36x^2+113x+72 = (4x+9)(9x+8) Given: 36x^2+113x+72 Use an AC method: Look for a pair of factors of AC=36*72=2^5*3^4 with sum B=113 Note that 113 is not divisible by 2 or 3 Hence all of the prime factors 2 must be in one of the pair and all of the factors 3 must be in one of the pair Try: 2^5+3^4=32+81 = 113" " Yes! So use the pair 2^5=32 and 3^4=81 to split the middle term, and factor by
|
|
|