Answer:
[tex]x=1, \quad x=\dfrac{3}{2}, \quad x=2[/tex]
Step-by-step explanation:
Factor Theorem
If f(x) is a polynomial, and f(a) = 0, then (x – a) is a factor of f(x).
If the coefficients in a polynomial add up to 0, then (x - 1) is a factor.
Given polynomial function:
[tex]f(x)=2x^3-9x^2+13x-6[/tex]
Sum the coefficients:
[tex]\implies 2-9+13-6=0[/tex]
As the sum of the coefficients is zero, (x - 1) is a factor.
Therefore:
[tex]\implies f(x)=(x-1)(ax^2+bx+c)[/tex]
Expand the brackets:
[tex]\implies f(x)=ax^3+bx^2+cx-ax^2-bx-c[/tex]
[tex]\implies f(x)=ax^3+(b-a)x^2+(c-b)x-c[/tex]
Compare the coefficients with the given polynomial.
As the coefficient of the leading term of the polynomial is 2:
[tex]\implies a=2[/tex]
As the coefficient of the term in x² is -9:
[tex]\implies b-a=-9[/tex]
[tex]\implies b-2=-9[/tex]
[tex]\implies b=-7[/tex]
As the constant of the given polynomial is -6:
[tex]\implies c=6[/tex]
Substitute the found values of a, b and c into the factored form of the polynomial:
[tex]\implies f(x)=(x-1)(2x^2-7x+6)[/tex]
Factor the quadratic:
[tex]\implies 2x^2-7x+6[/tex]
[tex]\implies 2x^2-4x-3x+6[/tex]
[tex]\implies 2x(x-2)-3(x-2)[/tex]
[tex]\implies (2x-3)(x-2)[/tex]
Therefore, the fully factored polynomial is:
[tex]\implies f(x)=(x-1)(2x-3)(x-2)[/tex]
To find the zeros, set the function to zero:
[tex]\implies f(x)=0[/tex]
[tex]\implies (x-1)(2x-3)(x-2)=0[/tex]
Apply the zero-product property:
[tex]\implies x-1=0 \implies x=1[/tex]
[tex]\implies 2x-3=0 \implies x=\dfrac{3}{2}[/tex]
[tex]\implies x-2=0 \implies x=2[/tex]
Therefore, the real zeros of the given polynomial are 1, ³/₂ and 2.
