Answer:
Step-by-step explanation:
Find the zero's of f(x)
Answer:
[tex]x=4, \quad x=1+i,\quad x=1-i[/tex]
(one real zero and 2 complex zeros)
Step-by-step explanation:
Given polynomial:
[tex]f(x)= 2x^3-12x^2+20x-16[/tex]
According to the Factor Theorem, if f(4) = 0 then (x - 4) is a factor of the given polynomial.
[tex]\implies f(x)=(x-4)(ax^2+bx+c)[/tex]
Expand the brackets:
[tex]\implies f(x)=ax^3+bx^2+cx-4ax^2-4bx-4c[/tex]
[tex]\implies f(x)=ax^3+(b-4a)x^2+(c-4b)x-4c[/tex]
Compare coefficients with the given polynomial:
[tex]\implies a=2[/tex]
[tex]\implies -4c=-16 \implies c=4[/tex]
[tex]\implies (b-4a)=-12 \implies b-8=-12 \implies b=-4[/tex]
Substitute the found values of a, b and c:
[tex]\implies f(x)=(x-4)(2x^2-4x+4)[/tex]
To find the zeros of f(x), set the function to zero and solve for x:
[tex]\implies (x-4)(2x^2-4x+4)=0[/tex]
[tex]\textsf{Therefore},\: (x - 4) = 0\: \textsf{ and }\: (2x^2-4x+4)=0[/tex]
[tex]\textsf{To solve}\: (2x^2-4x+4)=0\: \textsf{ use the quadratic formula}[/tex]
Quadratic Formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
Therefore:
[tex]\implies x=\dfrac{-(-4) \pm \sqrt{(-4)^2-4(2)(4)} }{2(2)}[/tex]
[tex]\implies x=\dfrac{4 \pm \sqrt{-16}}{4}[/tex]
[tex]\implies x=\dfrac{4 \pm \sqrt{16 \cdot -1}}{4}[/tex]
[tex]\implies x=\dfrac{4 \pm \sqrt{16}\sqrt{-1}}{4}[/tex]
[tex]\implies x=\dfrac{4 \pm 4i}{4}[/tex]
[tex]\implies x=1 \pm i[/tex]
Therefore, the zeros of the function are:
[tex]x=4, \quad x=1+i,\quad x=1-i[/tex]
(one real zero and 2 complex zeros)
