What is the horizontal intercept of f(x)=6x^3-19x^2+16x-4?

Question

What is the horizontal intercept of

f(x)=6x^3-19x^2+16x-4

Please show proof and will vote best answer ASAP.

Answer 1

f(x)=6x^3-19x^2+16x-4  = 0 gives  horizontal intercept

1. Obtain one factor (x -a ) by trial and error

2. . Factorise ax3 + bx2 + cx + d = 0 as (x - a) (hx2 + kx + s) = 0

3. Solve the quadratic expression for other roots.

By inspection, f(x) if of the 3rd degree, we would expect it to have 3 linear factors at most so that

f(x) = (x + a) (x + b) (x + c)

where a, b and c can be both positive and negative numbers. Also, by multiplying the last term of each factor, a x b x c numerically equals 4

so (x+1) ,(x-1), (x+2), (x-2)  (x-4), (x+4) can be factors

f(1)  = 6 -19+16-4    = -1  

f(-1) = -6-19-16-4

f(2)  = 6 2^3  -19 2^2 +16 * 2  -4 = 48 - 76 +32 -4  = 0

so x-2 is a factor

6x^3  - 19x^2 + 16x  -4   = (x-2)(ax^2+bx+c)

equating coeff x^3  we get  a = 6

equating const                   -4  = -2c    c = 2

equating coeff of x         16 = c-2b  = b   = -7

check  coeff of x^2     -19  = -2a+b  -2*6 -7 = -19 (correct)

so the equation is

6x^3  - 19x^2 + 16x  -4   = (x-2)(6x^2  -7x  +2)

solve the quadratic equation

                   -b + sqrt(b^2 - 4ac) / 2a and -b - sqrt(b^2 - 4ac) / 2a are the roots

                7 + sqrt( 49 - 4*2*6) / 2 *6    7  - sqrt( 49 - 4*2*6) / 2 *6

                8 / 12  or 6/12

  roots are   2/3    or 1/2

horizontal intercept   2  ,   2/3  . 1/2 subsitute and check