Question

If f(x)= 2x^3-12x^2+20x-16 and f(4)= 0, then find all the zeros of f(x) algebraically

Answer 1

`x=4, \quad x=1+i,\quad x=1-i`

(one real zero and 2 complex zeros)

Explanation:

Given polynomial:

`f(x)= 2x^3-12x^2+20x-16`

According to the Factor Theorem, if f(4) = 0 then (x - 4) is a factor of the given polynomial.

`\implies f(x)=(x-4)(ax^2+bx+c)`

Expand the brackets:

`\implies f(x)=ax^3+bx^2+cx-4ax^2-4bx-4c`

`\implies f(x)=ax^3+(b-4a)x^2+(c-4b)x-4c`

Compare coefficients with the given polynomial:

`\implies a=2`

`\implies -4c=-16 \implies c=4`

`\implies (b-4a)=-12 \implies b-8=-12 \implies b=-4`

Substitute the found values of a, b and c:

`\implies f(x)=(x-4)(2x^2-4x+4)`

To find the zeros of f(x), set the function to zero and solve for x:

`\implies (x-4)(2x^2-4x+4)=0`

`\textsf{Therefore},\: (x - 4) = 0\: \textsf{ and }\: (2x^2-4x+4)=0`

`\textsf{To solve}\: (2x^2-4x+4)=0\: \textsf{ use the quadratic formula}`

Quadratic Formula

`x=(-b \pm √(b^2-4ac) )/(2a)\quad\textsf{when }\:ax^2+bx+c=0`

Therefore:

`\implies x=(-(-4) \pm √((-4)^2-4(2)(4)) )/(2(2))`

`\implies x=(4 \pm √(-16))/(4)`

`\implies x=(4 \pm √(16 \cdot -1))/(4)`

`\implies x=(4 \pm √(16)√(-1))/(4)`

`\implies x=(4 \pm 4i)/(4)`

`\implies x=1 \pm i`

Therefore, the zeros of the function are:

`x=4, \quad x=1+i,\quad x=1-i`

(one real zero and 2 complex zeros)

Answer 2

• x = 4 is the only real zero

Explanation:

Find the zero's of f(x)

• 2x³ - 12x² + 20x - 16 = 0
• x³ - 6x² + 10x - 8 = 0
• x³ - 4x² - 2x² + 8x + 2x - 8 = 0
• x²(x - 4) - 2x(x - 4) + 2(x - 4) = 0
• (x - 4)(x² - 2x + 2) = 0
• (x - 4)(x² - 2x + 1 + 1) = 0
• (x - 4)[(x - 1)² + 1] = 0

• x - 4 = 0 ⇒ x = 4, we already know this
• (x - 1)² + 1 = 0 ⇒ (x - 1)² = - 1, no real solution as the square is never negative