How do I find all zeros of the function [math]f(x) =(x+3)^2?[/math]
Zero(es) of a function f(x) means the value(s) of x at which f(x)=0.So put (x+3)^2=0Thus we'll have x+3=0 (as square root of 0 is 0 only)Thus x=-3(-3+3)^2=(0)^2=0.So here f(x) has only one zero i.e -3
Find the zeros of the function.f(x) = 9x2 - 27x + 20?
Factor: (3x - 5)(3x - 4) = 0 So either 3x - 5 = 0; 3x = 5; x = 5/3 or (do the same for 3x - 4)
F(x)=x^2-9x+14/4x find the zeros of the function?
(x-7)(x-2)/4x = 0 (x-7)/4x = 0 x-7 = 0 x = 7 (x-2)/4x = 0 x-2 = 0 x = 2
Find all the zeros of a function: f(x)= 8x^3-64x^2+160x-128?
let's fin the factors between the leading coefficient which is 8 and the constant term -128 +/-1 , +/-2 , +/-4 , +/-8 , +/- 16 , +/- 32 , +/- 64 , +/- 128 , +/- 1/2 , +/- 1/8 , ... Now, we are going to try each number plugging in the original equation and the one makes it zero as the following f(2) = 8 * (2)^3 - 64 * (2)^2 + 160 * 2 - 128 = 0 (good) x = 2 ===> x - 2 = 0 let's do long division: . . . . . 8x^2 - 48x + 64 . . . . . ___________________ x - 2 I 8x^3 - 64x^2 + 160x - 128 . . . . .- . . . . .8x^3 - 16x^2 . . . . . ____________________ . . . . . 0 - 48x^2 + 160x - 128 . . . . . - . . . . . . . -48x^2 + 96x . . . . . ____________________ . . . . . .. . .0 + 64x - 128 . . . . . - . . . . . . . .. . . .64x - 128 . . . . . ____________________ . . . . . .. 0 + 0 f(x) = 8x^3 - 64x^2 + 160x - 128 = (x - 2) * (8x^2 - 48x + 64) f(x) = 8x^3 - 64x^2 + 160x - 128 = (x - 2) * 8 * (x^2 - 6x + 8) f(x) = 8x^3 - 64x^2 + 160x - 128 = (x - 2) * 8 * (x - 4)(x - 2) f(x) = 8x^3 - 64x^2 + 160x - 128 = (x - 2)^2 * 8 * (x - 4) x = 2 & 4 <=== roots
How do you find all rational zeros for f(x) =x^3-2x^2-5x+6?
As the coefficient of x^3 is 1, rational zeros are integers. The numerical term has only eight divisors, 1, -1, 6, -6, 2, -2, 3 and -3. So, if there are any rational roots, one of them must be one. Try them. 1^3 - 2(1^2) - 5(-1) + 6 = 0.Yes, you have found a zero straightaway.Now factorise: (x-1)(x^2 - Ax - 6) = x^3 - (A+1)x^2 + (A-6)x + 6 so A must be 1.Therefore x = 1 or x^2 - x - 6 = 0. The latter is a quadratic with discriminant 1+24 = 25. As this is a perfect square there must be two integer solutions, so factorise it. You should find (x - 3)(x + 2) = 0 so solutions are 1, 3 and -2.They are all rational.
F(x)=5x-3 find the zeros?
f(x)=5x-3 5x - 3 = 0 5x = 3 x = 3/5 answer//
How do you find the zeros of f(x) = x^5-2x^4+8x^2-13x+6?
Find first the factorization indicated by Eric D. Then use the general result for cubic equations of the form A x^3 + B x^2 + C x + D = 0. You can also use only one of the first roots in the factorization, then use the general result for quartic equations of the form A x^4 + B x^2 + C x^2 + D x + E = 0. You can find both formulas in any good standard algebra book.
Find the zeros of f(x) = 9x4 - 37x2 + 4?
f(x) = 9x4 - 37x2 + 4 0 = 9x4 - 37x2 + 4 0= (9x^2-1)(x^2-4) 0=(3x-1)(3x+1)(x-2)(x+2) x= 1/3 x= -1/3 x=2 x=-2