Question

According to the Fundamental Theorem of Algebra, the graph of f(x) = x2 - 4x + 3, has roots. From the graph we can see that it has zeros.

Answer 1

Answer:

The graph $f(x)=x^2-4x+3$  has two zeros namely 3 and 1.

Step-by-step explanation:

Consider the given equation of graph $f(x)=x^2-4x+3$

According to the Fundamental Theorem of Algebra

For a given polynomial of degree n can have a maximum of n roots.

Thus, for the given equation $f(x)=x^2-4x+3$  the degree of polynomial is 2 , thus the function can have maximum of 2 roots.

We know at roots the value of function is 0 that is f(x) = 0,

Substitute f(x) = 0 , we get, $f(x)=x^2-4x+3=0$

This is a quadratic equation, $x^2-4x+3=0$

We first solve it manually and then check by plotting graph.

Quadratic equation can be solved using middle term splitting method,

here, -4x can be written as -x-3x,

$x^2-4x+3=0 \Rightarrow x^2-x-3x+3=0$

$\Rightarrow x(x-1)-3(x-1)=0$

$\Rightarrow (x-3)(x-1)=0$

Using zero product property, $a\cdot b=0 \Rightarrow a=0\ or \ b=0$

$\Rightarrow (x-3)=0$ or $\Rightarrow (x-1)=0$

$\Rightarrow x=3$ or $\Rightarrow x=1$

Thus, the two zero of f(x) are 3 and 1.

We can also see on graph attached below that the graph $f(x)=x^2-4x+3$  has two zeros  namely 3 and 1.