# How do you find the zeros of f(x)=5x^2-25x+30?

Nov 12, 2017

See a solution process below:

#### Explanation:

We can factor this function as:

$f \left(x\right) = \left(5 x - 15\right) \left(x - 2\right)$

To find the zeros we can solve each term on the right side of the function for $0$:

Solution 1:

$5 x - 15 = 0$

$5 x - 15 + \textcolor{red}{15} = 0 + \textcolor{red}{15}$

$5 x - 0 = 15$

$5 x = 15$

$\frac{5 x}{\textcolor{red}{5}} = \frac{15}{\textcolor{red}{5}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} x}{\cancel{\textcolor{red}{5}}} = 3$

$x = 3$

Solution 2:

$x - 2 = 0$

$x - 2 + \textcolor{red}{2} = 0 + \textcolor{red}{2}$

$x - 0 = 2$

$x = 2$

The Solutions Are: $x = \left\{2 , 3\right\}$

Nov 12, 2017

$x = 2 , x = 3$

#### Explanation:

$\text{to calculate the zeros set } f \left(x\right) = 0$

$\Rightarrow 5 {x}^{2} - 25 x + 30 = 0 \leftarrow \textcolor{b l u e}{\text{factorise to solve}}$

$\Rightarrow 5 \left({x}^{2} - 5 x + 6\right) = 0$

$\text{the factors of + 6 which sum to - 5 are -2 and - 3}$

$\Rightarrow 5 \left(x - 2\right) \left(x - 3\right) = 0$

$\text{equate each factor to zero and solve for x}$

$x - 2 = 0 \Rightarrow x = 2$

$x - 3 = 0 \Rightarrow x = 3$

$\text{the zeros are } x = 2 , x = 3$
graph{(y-x^2+5x-6)((x-2)^2+y^2-0.07)((x-3)^2+y^2-0.07)=0 [-10, 10, -5, 5]}