# What are the zeroes of f(x) = x^{2} - 6x + 8?

x = -4, 2

x = -4, -2

x = 4, 2

x = 4, -2

**Solution:**

The zeroes of the polynomial make the values of the whole polynomial equal to zero.

Let us factorise the polynomial to find the value of x by splitting the middle term.

**Step 1: **

Identify the values of a, b and c.

In the above equation, a is coefficient of x^{2 }= 1,

b is the coefficient of x = - 6 and c is the constant term = 8.

**Step 2:**

Multiply a and c and find the factors that add up to b.

1 × (8) = 8

⇒ - 4 and - 2 are the factors that add up to b.

**Step 3:**

Split bx into two terms.

x^{2} - 4x - 2x + 8 = 0

**Step 4: **

Take out the common factors by grouping.

x(x - 4) - 2 (x - 4) = 0

(x - 4) (x - 2) = 0

**By putting the factors equal to zero we get two values of x**

x - 2 = 0 and x - 4 = 0

x = 2 and x = 4

**Thus, the two values that satisfy the equation are 2 and 4.**

## What are the zeroes of f(x) = x^{2} - 6x + 8? x = - 4, 2 x = - 4, - 2 x = 4, 2 x = 4, - 2

**Summary:**

The zeroes of the equation f(x) = x^{2} - 6x + 8 are x = 2, 4 which satisfies the equation.