How do you factor [math]x^3-6x^2+11x-6=0[/math]?
Given (x^3)-6(x^2)+11x-6=0x^3-3x^2-3x^2+11x-6=0x^2(x-3)-(3x^2-11x+6)=0x^2(x-3)-(3x^2-9x-2x+6)=0x^2(x-3)-(3x(x-3)-2(x-3))=0x^2(x-3)-[(x-3)(3x-2)]=0(x-3)(x^2-3x+2)=0(x-3)(x^2-2x-x+2)=0(x-3)(x(x-2)-1(x-2))=0(x-3)(x-2)(x-1)=0Online Math Tutor
How to factor 3x² -9x - 30 ?
You should take out the common factor of 3 first so it becomes 3(x^2-3x-10) Then you need two number that multiply to give 10 and can be subtracted to give -3. These are 2 and -5 The answer is therefore 3(x+2)(x-5)
How do you factor this: 36x^2-120x+100?
Hi, 36x² - 120x + 100 = 4(9x² - 30x + 25) = 4(3x - 5)² <==ANSWER I hope that helps!! :-)
Solve for x using the factoring method: -30 = 19x - 5x² and 9x² - 30x + 25 = 0 PLEASE CAN U SHOW ME HOW TO DO
-30 = 19x-5x^2 or 5x^2-19x-30 = 0 it is an quardic equation, having two roots or solution 5x^2-25x+6x-30 = 0 5x(x-5)+6(x-5)= 0 (x-5)(5x+6) = 0 either (x-5) = 0 or (5x+6) = 0 if (x-5) = 0 x = 5 or (5x+6) = 0 x = -6/5 and 9x^2-30x+25 = 0 9x^2-15x-15x+25 = 0 3x(3x-3)-5(3x-5) = 0 (3x-5)(3x-3) = 0 on solving these x = 5/3 x = 1
How do I factor this equation using decomposition? 6x^2+11x+3
Thank you for the A2A!So, decomposition is a way of factoring that allows us to break apart complex trinomials and make them easier to factor.The first step in decomposition would be, in the form ax^2 +bx +c, find which two numbers add to b, and multiply to a*c.In the case of 6x^2 +11x + 3, we need two numbers that add to 11 and multiply to 18.Quickly cycling through the factors of 18 leaves us with 9 and 2 (multiply to 18, add to 11).From there, we break the middle term into these two parts:6x^2 +11x + 3 becomes:6x^2 + 9x +2x + 3Then, we can factor each half by getting the gcf of the first two terms, and the gcf of the second two terms. 6x^2 + 9x +2x + 3becomes:3x (2x+3) + 1 (2x+3)Which then can be factored again by taking out the 2x+3, giving us a final answer of (3x+1)(2x+3)I really hope this helps! Please let me know if you have any questions!Conner D
Factoring help?
1. x^2+2xy+y^2+4x+4y+4 2. a^2+2ab+b^2+6a+6b+9 3. 16+8r+8s+r^2+2rs+s^2 4. 25+10m+10n+m^2+2mn+n^2 5. 6x^3+2x^2y+21x^2+7xy+15x+5y 6.20a^3+5a^2b-68a^2-17ab+24a+6b 7. 8a^3+12a^2b-24a^2-36ab+18a+27b 8. 9x^3+6x^2y-45x^2-30xy+36x+24y 9.6x^3-2x^2-21x^2y+7xy-45xy^2+15y^2 10. 15x^2+30xy+15y^2-7x-7y-2 11. 8a^2+16ab+8b^2+18a+18b+9 12. 28c^2+56cd+28d^2-c-d-2
Factorize x³-6x² +3x+ 10?
Apply trial and error to the equation by substituting x as any random integer to the equation [math] x^3 - 6x^2 + 3x + 10 = 0 [/math]. If the equation gives a result of 0, then (x - root) will be one factor of the equation. In that case, consider x = 2, so [math] 2^3 - 6(2)^2 + 3 (2) + 10 = 0 [/math] [math] 8 - 24 + 6 + 10 = 0 [/math] Hence, (x - 2) is one of the roots of the equation. Now, apply long division, and accordingly we will obtain another factor as [math] x^2 - 4x - 5 = (x - 5) (x + 1) [/math]. Hence, [math] x^3 - 6x^2 + 3x + 10 = (x + 1) (x - 2) (x - 5) [/math].
Algebra Question: Completely factor 5xsquared -15x-90 which of the following is a factor?
5x^2-15x-90 5(x^2-3x-18) 5(x-6)(x+3)
Without actual division, can you prove that ‘2x^4-6x^3+3x^2+3x-2’ is exactly divisible by ‘x^2-3x+2’?
If P(x)=2x^4–6x^3+3x^2+3x-2 is divisible byg(x)=x^2–3x+2=(x-2)(x-1) thenP(1)=0 and P(2)=0P(1)=2–6+3+3–2=8–8=)P(2)=2×16–6×8+3×4+3×2–2=32–48+12+6–2=50–50=0Therefore P(x) is divisible by g(x).