Homework help, please. Factoring, Algebra 2?
Write the correct factorization for 81x^2 - 4 = 0. Then find the solution(s). 81x^2 - 4 = 0 (9x + 2)(9x + 2) = 0 9x + 2 = 0 or 9x + 2 = 0 x = - 2/9 81x^2 - 4 = 0 (9x + 2)(9x - 2) = 0 9x + 2 = 0 or 9x - 2 = 0 x = 2/9, -2/9 81x^2 - 4 = 0 9x ( x - 4) = 0 9x = 0 or x - 4 = 0 x = 0, 4 81x^2 - 4 = 0 (9x - 2)(9x - 2) = 0 9x - 2 = 0 or 9x - 2 = 0 x = 2/9 Which one guys?
Algebra 2 Homework, Factoring?
there's no difference, you just have to factor the other #s in as well so: 6x^2+49x-45 (6x-5)(x+9) and on this next one, you want to make it so that it equals zero, then put it in order and factor so: -53x+95= -5x^2-9 +5x^2+9= +5x^2+9 5x^2-53x+104 (5x-13)(x+8) and this one is done the same way: 8x^2-55x= -42 +42 +42 8x^2-55x+42 (8x-7)(x-6) and finally: 3x^2+11x-20 (3x-4)(x+5)
Algebra factoring homework help?
Solve the equation: 0 = -16t^2 + 8t + 80 Divide by -8: 0 = 2t^2 - t - 10 Factor the quadratic: 0 = (t + 2) (2t - 5) Use the zero product principle and discard the negative solution: 2t - 5 = 0 2t = 5 t = 5/2 seconds
Algebra II homework help???? Factoring.?
2x^3 - 13x^2 - 48x + 27 = 2x^3 -- 18x^2 + 5x^2 -- 45x -- 3x + 27 = 2x^2(x -- 9) + 5x(x -- 9) --3(x -- 9) = (x -- 9)(2x^2 + 5x -- 3) = (x -- 9)(x + 3)(2x -- 1)
Algebra 2 Homework Help! Factoring completely!?
Factor completely. a² - 3ab - 130b² = (a + 10b)(a - 13b) 8y³ + 24y² - 7y - 21 = (8y³ + 24y²) - (7y + 21) 8y³ + 24y² - 7y - 21 = 8y²(y + 3) - 7(y + 3) 8y³ + 24y² - 7y - 21 = (y + 3)(8y² - 7) 32³ - 108d³ = 32768 - 108d³ 32³ - 108d³ = 4(8192) - 4(27d³) 32³ - 108d³ = 4(8192 - 27d³) xy + yz - xw - wz = (xy + yz) - (xw + wz) xy + yz - xw - wz = y(x + z) - w(x + z) xy + yz - xw - wz = (x + z)(y - w) x² + 12x + 36 - 25y² = (x² + 12x + 36) - (25y²) x² + 12x + 36 - 25y² = (x + 6)² - (5y)² x² + 12x + 36 - 25y² = [(x + 6) + (5y)][(x + 6) - (5y)] x² + 12x + 36 - 25y² = (x + 6 + 5y)(x + 6 - 5y) x² + 4x + 4 - y² y²x^4 - 16y^6 Solve. 12x² + 11x - 15 = 0
HOMEWORK HELP! PLEASE! ALGEBRA! FACTORING WITH VARIABLES!?
How do I factor: 15a - 25b + 20 'll give you an example of one without variables 11^2 - 6(11) + 5(11) Greatest Common Factor is 11 so you write: 11( 11- 6 +5) You solve what's in the parenthesis.... 11 (10) And the answer is 110 How do I do that with this problem? 15a - 25b + 20
Factoring by Grouping Algebra homework help?
im having a little trouble with my algebra homework. im a little lost and whenever i ask my teacher for help, i get even more confused. if anyone could help me with these problems it would be great! Factor. Check by multiplying the factors. 1) e(f - g ) - 4 ( f - g) 2) 7(m - n) + p (n - m) 3) 2x(x - y) + y(y - x) 4) a(a - b) + 4b(a - b) - a(a - b) 5) (s^2 - 2ps + 2s) - (2s - 4p + 4) if someone could please help me it would be great!! i dont get this :/
Algebra Factoring! HELP!?
10. Suppose that we have a 3 ft. × 3 ft. piece of paper, and we cut out a square piece so that the area of the remaining piece is 5 feet. What is the length of a side of the square? A. square root of 2 B. square root of 5 C. 4 ft. D. 2 ft. Solve the equation x2 – 2/3 x + 1/9 = 0. A. { 1/3, -1/3} B. 1/9 C. {1/9, -1/9} D.1/3 Thanks sooo much!
Algebra I Factoring Help?
I just learned this a while back in math. I will try and help with two and maybe teach you a little. Sorry if none of two work out well. Anyways, so 2Q^2 -11Q - 21. Step 1: Just start with two pairs of brackets: ( )( ) Step 2: Look at the the term (assuming you have learned about polynomials) in the position of ax^2. In this case, the term is 2Q^2. In each bracket, write the square root of 2Q^2. NOTE: You DO NOT find the square root of the coefficient (2). So that leaves only one possible way: (2Q )(Q ) Step 3: Look at the term in the position of +c. In this case, the term is -21. Find a pair of numbers that equals -21 when they are multiplied (i.e -7 multiplied by 3). Place them into the brackets. Position DOES matter but you don't know which position just yet. So just place them with their signs for now. (2Q -7)(Q +3). Now multiply the binomials and see if you get the result 2Q^2 -11 -21. Your answer should be: 2Q^2 -Q -21. So that is not the answer. Now, exchange the positions of -7 and 3 and try it. Your answer should be: 2Q^2 -11Q -21. If the product of the two binomials does not equal the original trinomial, then try another pair of numbers. Step 4: If after trying all the other possible pairs of numbers, you still cannot get the product of the binomials to equal the original trinomial, then you can safely say that it is impossible to factor the trinomial that you were given. Lets try another one shall we? 21V^2 -70V +49. Step 1: ( )( ) Step 2: The term in the position of ax^2 is 21V^2. So fill your brackets accordingly: (7V )(3V ). Step 3: Fill in -7 into both brackets (you know its negative 7 and not positive 7 because the coefficient in position of +bx is a negative number) like this: (7V -7)(3V -7). Find the product of the two binomials that we came up with. Your answer should be 21V^2 -70V +49. Congratulations! You now have the knowledge to do the rest! At least, I hope.