How do you factor this: 36m 4 - 49n 4?
36m^4 - 49n^4 = (6m^2 - 7n^2)(6m^2 + 7n^2) it can be factored futher but not over the integers Use quadratic equation to factor further
How do you factor 9x^2+42x-15?
You can factor this using the AC Method. A = 9 B = 42 C = -15 ac = (9)(-15) = -135 So, you replace the middle term with 2 integers that have a product equal to ac and a sum equal to b. The integers are: 9x^2 - 3x + 45x - 15 Then factor out the common terms. = 9x^2 - 3x + 45x - 15 = 3x (3x -1) + 15 (3x - 1) Get the common terms. = 3x (3x -1) + 15 (3x - 1) = (3x - 1) (3x + 15) You can still factor out (3x + 15). So, = (3x - 1) (3x + 15) = (3x - 1) [(x + 5)3] Therefore, the answer is (3x - 1) [(x + 5)3]
How do you factor this 9x^3+30x^2-24x?
Hi, For this question, notice that there's a common factor in each term of x. Now, also notice that the coefficients of each term also have a common factor of 3. If you have trouble with this, take each term and divide it by 3x and see if you're able to do that. If you can do it, then it's a common factor! :=) Now, let's take these common factors out of the polynomial to get: 3x ( 3x^2 + 10x - 8 ) Now, when you're put with this situation now, you need to question whether the Quadratic can or can not be factored. You can try this through the Quadratic Formula and will eventually get: x = -5 +/- 7 divided by 3 Therefore, x equals: x = -4 and x = 2/3 Now, we can say that the fully factored form of your expression is: 3x ( x + 4 ) ( x - 2/3) <=== FINAL ANSWER BTW, If you want to avoid the fraction in the factored form, try multiplying the two parenthetical quantities by 3 to get: 3x ( 3x + 12 ) ( 3x - 2 ) <==== FINAL ANSWER I hope that helps you out! Please let me know if you have any other questions!
How do I factor 4x^2+12x+9?
4x^2+12x+9So for factors, u have to split 12x in such a way that some common factors can come out.4x^2+12x+9= 4x^2+6x+6x+9= 2x(2x+3)+3(2x+3)= (2x+3)(2x+3)To find the value of X ,(2x+3)(2x+3)=02x+3=02x=-3x=-3/2 answer
6n-4n^2 how do you factor this?
2n is common, so you can factor it out. 2n(3-2n)
How do you factor this...6x^2-26x-20??
The best way to get good at these is just practice Factor out a 2 first 2(3x^2-13x-10) 2(3x+2)(x-5) When I factor this, I though "What two numbers multiply to 10 and add up to -13? But the trick is that 3. So I would list the factors of 10 1,2, 5, 10 So it's either 1 and 10 or 2 and 5. It's usually not 1 and the number (just a piece of advice). Let's try 2 and 5.Can we multiply one of these number be 3, and then subtract or add the other number? Let's try 3*-2-5. That's only -11, not -13. 3*-5+2. That's -13. The 3 and -5 must be in opposite parentheses. So then it would be (3x+2)(x-5)
How do you factor this a(b-c) ^3+b(c-a) ^3+c(a-b) ^3?
Call the expression [math]E[/math]. After noting that [math]E[/math] is cyclic, conclude that it can be factored into cyclic factors but count the degree of the expressions as you factor.Since [math]a=b[/math] make [math]E=0[/math], [math](a-b)[/math] must be a factor. Being cyclic, so must [math](b-c)[/math] and [math](c-a)[/math]. So far we have[math]E=k(a-b)(b-c)(c-a)[/math].As [math]E[/math] is of degree [math]4[/math] and we have accounted for three of them, the remaining factor must be a cyclic factor of degree one. The only such factor is [math]a+b+c[/math]. Observing that [math]E[/math] contains a term spelled [math]ab^3[/math] and [math](a+b+c)(a-b)(b-c)(c-a)[/math] contains [math]-ab^3,[/math] we term one factor around and say[math]E=(a+b+c)(b-a)(b-c)(a-c).[/math]Please: When I factor for fun or profit, I really don’t write all those words
How can you factor this a^2(b-c) +b^2(c-a) +c^2(a-b)?
This is a classic.a^2(b - c) + b^2(c-a) + c^2(a - b)= a^2b - a^2c + b^2c - b^2a + c^2a-c^2b=a^2b - b^2a - a^2c + b^2c + c^2a -c^2b= ab(a - b) - c(a^2 - b^2) +c^2(a - b)=ab(a - b) -c(a - b)(a + b) +c^2(a - b)=(a - b)(ab - c(a + b) +c^2)=(a - b)(ab - cb -ca + c^2)= (a - b)(b(a-c) -c(a-c))=(a-b)(b-c)(a-c)
How do you factor this?
We're doing the LCM of polynomials and you have to factor it first before you can find the LCM. I've tried and tried and tried and I can't figure this out! x³ + 4x² - x - 4 22x² + 66x - 220
How do you factor this?
Find the factors of 10x^2: 10x & x, 5x & 2 Find the factors of -6y^2: -6y & y, 6y & -y, -3y & 2y, 3y & - 2y Now, find which combination gives a middle term of +11xy: (10x)(-6y) + (x)(y) = -60xy + xy = -59xy NO (10x)(6y) + (x)(-y) = 60xy - xy = 59xy NO (10x)(-y) + (x)(6y) = -20xy + 6xy = -14xy NO (10x)(y) + (x)(-6y) = 20xy - 6xy = 14xy NO (10x)(-3y) + (x)(2y) = -30xy + 2xy = -28xy NO (10x)(3y) + (x)(-2y) = 30xy - 2xy = 28xy NO (10x)(-2y) + (x)(3y) = -20xy + 3xy = -17xy NO (10x)(2y) + (x)(-3y) = 20xy - 3xy = 17xy NO (5x)(-6y) + (2x)(y) = -30xy + 2xy = -28xy NO (5x)(6y) + (2x)(-y) = 30xy - 2xy = 28xy NO (5x)(-3y) + (2x)(2y) = -15xy + 4xy = -11xy NO (5x)(3y) + (2x)(-2y) = 15xy - 4xy = 11xy YES!!! Now, we can write it in factored form: (5x - 2y)(2x + 3y) Check your answer using FOIL (first outer inner last): (5x)(2x) + (5x)(3y) + (2x)(-2y) + (-2y)(3y) 10x^2 + 15xy - 4xy - 6y^2 10x^2 + 11xy - 6y^2