How do i solve these quadratic equations by factoring?
The first one you just have to realize it's a difference of perfect squares. (x+7)(x-7) Second one you can factor out a 3 and an x right? 3x(x^2 - 4) Which also happens to be a difference of perfect squares. Factor 1 more time. Third you gotta subtract first, then factor out a 2. You get 2(6x^2 + 7x - 3). That one factors out too. 3x-1 and 2x+3 Fourth, factor out a negative and an x. -x(x^2 + 22x - 121) That's a square of a binomial. Factor it out. Last, subtract and factor. (x-5) (x+1) Factoring is more of a trial and error type of math question. Ask your teacher for more help.
Factor: x^2-4x-5 ???????
(x - 5) (x + 1)
Ples help Factor and solving show you to do?
36x^2 = -65x + 36 One way to do this without thinking too much is by completing the square. Add 65x to both sides : 36x^2 + 65x = 36 Divide through by 36 : x^2 + (65/36)x = 1 Take half of 65/36 (= 65/72) and square it (= 65^2/72^2). Add this to both sides : x^2 + (65/36)x + 65^2/72^2 = 1 + 65^2/72^2 Form a square on the LHS and sum the RHS : (x + 65/72)^2 = 9409/72^2 Take the square root of both sides : x + 65/72 = ± 97/72 Subtract 65/72 from both sides : x = 4/9 or -9/4 These are the two roots of 36x^2 + 65x - 36 = 0, so equivalently, this may be represented as : (x - 4/9)(x + 9/4) = 0 Multiply left-hand term by 9 and right-hand term by 4 : (9x - 4)(4x + 9) = 0 Thus, 36x^2 + 65x - 36 = 0 can be factored as (9x - 4)(4x + 9) = 0 and we already have the answers for x.
How can one solve the following by using the method of factorization: [math]\frac{4}{x-3}=\frac{5}{2x+3}[/math]where [math]x[/math] is not equal to [math]0[/math] or [math]-\frac{3}{2}?[/math]
4/x-3=5/2x+3Multiply both sides by (x-3)(2x+3)4(2x+3)=5(x-3)=8x+12=5x-15Subtract 12 from both sides8x+12 -12 =5x-15 -12= 8x=5x-27Subtract 5x from both sides.8x -5x =5x-27 -5x= 3x=-27x=-9
How do l factorise [math]x^3-3x^2-9x-5[/math] by using factor theorem?
x^3 - 3x^2 - 9x - 5p(x) = x^3 - 3x^2 - 9x - 5Factors of 5 = 1,5,-1,-5p(1) = 13 - 3 x (1)2 - 9 x (1) - 5 = 1 -3 - 9 - 5 = 1 - 17 = - 16 ≠ 0p(-1) = (-1)3 x 3 x (-1)2 - 9 x (-1) -5 = -1 - 3 + 9 - 5 = -9 + 9 = 0p(-1) =0∴ (x+1) is a factor of p(x) p(x) ÷ (x+1) x3 - 3x2 - 9x - 5 ÷ x + 1 = x2 - 4x - 5x2 - 4x - 5x2 - 5x + 1x - 5 = x ( x - 5) + 1 ( x - 5)= ( x - 5) ( x + 1Factor Theorem
Help solve 5x^2 - 24x - 5 = 5?
Please restate the question. x is irrational, so you cannot do this by factoring, only by completing the square or the quadratic equation. 5x^2 - 24x - 5 = 5 5x^2 - 24x - 10 = 0 If this were possible by factoring, you would be able to find two integers with a product of 50 and with a difference of 24. Factors of 50: 1, 50, difference = 49 2, 25, difference = 23 5, 10, difference = 5 If your question is 5x^2 - 23x - 5 = 5, then it is possible. 5x^2 - 23x - 5 = 5 5x^2 - 23x - 10 = 0 -> First you want right hand side to be 0 when solving by factorization, always. 5x^2 - 25x + 2x - 10 = 0 -> As I stated above, find two integers with the sum of the first-order coefficient, the middle term, and with a product equal to the product of the second-order coefficient and the constant term, two outside terms. Here, -25 + 2 = -23 and -25 * 2 = -50 = 5 * -10. 5x(x - 5) + 2(x - 5) = 0 -> Factor out like terms for both the first two terms and the last two terms. (5x + 2)(x - 5) = 0 -> Combine using ab + ac = a(b + c), here a = x - 5, b = 5x, and c = 2. 5x + 2 = 0 or x - 5 = 0 x = -2/5 or x = 5 Again, if your term is in fact -24 instead of -23, you will have to use complete the square or the quadratic formula, but it is not possible with factorization. EDIT: 5x^2 -24x - 5 = 0 Same thing. Find two numbers with a difference of 24 and a product of 25. 1 and 25 easily work. 5x^2 - 24x - 5 = 5x^2 - 25x + x - 5 = 5x(x-5) + (x-5) = (5x+1)(x-5) = 0 So x = -1/5 or x = 5