How to factorise 9x^2+4x-6?
because it doesn't factor evenly you have to use the quadratic formula.
How do you factorise ax² - bx² - 2bx + 2ax - 3a + 3b?
I advise students studying elementary algebra that any time you see an even number of terms and the goal is factoring, it is possible that you are expected to factor by grouping.First, we can add some parenthesis just so you can see the groups. Also, I’ll rearrange the middle group since it’s been intentionally obscured. The third group has also been intentionally obscured a bit, but if we imagine pulling the sign in front of the 3a term out of the grouping, we get this:ax² - bx² - 2bx + 2ax - 3a + 3b = (ax² - bx²) + (2ax-2bx) - (3a - 3b)Now, within each group, pull out the greatest common factorx²(a - b) + 2x(a-b) - 3(a-b)So now we have a sum of products, and within each piece of the sum, there is a common factor. So it can be pulled out. What is left is a simple quadratic. Good luck finishing it up.
How do I factorize: 25 (a+2b)^2 -36 (2a-5b) ^2?
I strive to save time when it comes to algebra…I just inputFactor [ 25 (a + 2 b)^2 - 36 (2 a - 5 b)^2 ]into my Mathematica app and I got-(7 a - 40 b) (17 a - 20 b)(you may use Computational Knowledge Engine (Wolfram Alpha) if you don't have Mathematica.But if you ask what the method of solution is (to do it by hand), then I’d suggest to expand everything out then factor by means of the long-division-of-polynomials algorithm - tedious, tedious, thanks God for symbolic computation programs such as Mathematica and the Wolfram site.
How do I factorise [math]2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4[/math]?
[math](x-y-z)^2=x^2+y^2+z^2-2(xy-yz+xz)[/math][math]x^2-y^2=(x-y)(x+y)[/math][math](x-y)^2=x^2-2xy+y^2[/math]This identities can be proved easliy by induction method. Hence by using this identities we get[math]2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4=-[a^4+b^4+c^4-2(a^2b^2-b^2c^2+c^2a^2)]+4b^2c^2[/math][math]=4b^2c^2-(a^2-b^2-c^2)^2[/math][math]=(2bc-a^2+b^2+c^2)(2bc+a^2-b^2-c^2)[/math][math]=[(b^2+2bc+c^2)-a^2][a^2-(b^2-2bc+c^2)][/math][math]=[(b+c)^2-a^2][a^2-(b-c)^2][/math][math]=(b+c-a)(b+c+a)(a+b-c)(a-b+c)[/math]
Factor: 3a - 3b + ax - bx?
3a - 3b + ax - bx = 3(a - b) + x (a - b) = (3 + x)(a - b)
How do you factor 2a^2+6a-ab-3b?
Well you know because its a quadratic that you'll have two binomials and because you have 2a² you can set up an initial multiplication (2a ± ?)(a ± ?) Now you just have to kinda think through it, Notice that both the terms with b are negative so you can assume that b is negative, you can assume (2a - b)(a ± ?) or (2a ± ?)(a - b) Here you might notice that the left one works better because right would result in -2ab which you don't have in the original. So now all you have to do is find the last term, and because there is only one term with one b you know that it's 3. (2a - b)(a + 3) Multiply it out to check your answer 2a² + 6a - ab - 3b Yep we were right. Davud's answer would be easier... factor by grouping. (2a² + 6a) + (-ab - 3b) Now just factor out the whatever you can to make the two binomials the same. 2a(a+3) - b(a+3) Then factor out the binomial (a+3)(2a-b) Sorry I forgot about grouping pretty much as soon as I learned it.
Please help me factorise!?
[2(3a+b) + 3(4a-3b)][2(3a+b) + 3(4a-3b)] (6a + 2b + 12a - 9b)(6a + 2b - 12a + 9b) (18a - 7b)(-6a + 11b) 4x^2 - 2(5x+3) 4x^2 - 10x - 6 4x^2 - 12x + 2x - 6 4x(x - 3) + 2(x - 3) (4x + 2)(x - 3) 8x^3 + 27y^3 - 6x^2 - 5xy + 6y^2 (2x + 3y)(4x^2 + 6xy + 9y^2) - (6x^2 + 5xy - 6y^2) (2x + 3y)(4x^2 + 6xy + 9y^2) - (2x + 3y)(3x - 2y) (2x + 3y)(4x^2 + 6xy + 9y^2 - 3x + 2y) 2(2m+1)^2 - 7(2m+1) + 6 2(2m+1)^2 - 3(2m+1) - 4(2m+1) + 6 (2m+1)[2(2m+1) - 3] - 2[2(2m+1) - 3] (2m + 1 - 2)(4m + 2 - 3) (2m - 1)(4m - 1)
How would you go about factoring 6b^2+ab-15a^2?
I know how to factor a polynomial that only has 1 variable, however, when 2 variables are put into the equation, I can find the answer but I am supposed to show all my work. Can anyone show me the steps to get from the question above to the answer??
How Would you Work Out This? Factorise fully 3a^3b+12a^2b^2+9a^5b^3?
3a^3b + 12a^2b^2 + 9a^5b^3 = 3(a^3b + 4a^2b^2 + 3a^5b^3) = 3a²b(a + 4b + 3a³b²) You sure that last term isn't supposed to be 9ab³? If it is, the expression factors to: 3a³b + 12a²b² + 9ab^3 = 3ab(a² + 4ab + 3b²) = 3ab(a+b)(a+3b)
How do I factorize [math] 2a^{2} + 5ab + 2b^{2} [/math]?
This can be solved by using middle term splitting…2a^2 + 5ab + 2b^2=>2a^2 + 4ab + ab + 2b^2=>2a(a+2b) + b(a+2b)=>(2a+b) (a+2b)