# How do you find the zeroes of f(x) = (3x-5)(2x+7)?

Jun 5, 2018

$x = - \frac{7}{2} , x = \frac{5}{3}$

#### Explanation:

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

$\left(3 x - 5\right) \left(2 x + 7\right) = 0$

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

$2 x + 7 = 0 \Rightarrow x = - \frac{7}{2}$

$3 x - 5 = 0 \Rightarrow x = \frac{5}{3}$

Jun 5, 2018

Zeros are $x = \frac{5}{3}$ and $x = - \frac{7}{2}$

#### Explanation:

Zeros, Roots and Solutions to a polynomial are all the same thing, to find the zeros we first set $f \left(x\right) = 0$ and then solve for x:

$f \left(x\right) = \left(3 x - 5\right) \left(2 x + 7\right)$

$0 = \left(3 x - 5\right) \left(2 x + 7\right)$

The zero product rule states that if:

$a \cdot b = 0$ then $a = 0$ or $b = 0$

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

$3 x - 5 = 0$

$3 x = 5$

$x = \frac{5}{3}$

or

$\left(2 x + 7\right) = 0$

$2 x + 7 = 0$

$2 x = - 7$

$x = - \frac{7}{2}$

Jun 5, 2018

f(x)=0 when $x = \frac{5}{3}$ and $x = - 3 \frac{1}{2}$

#### Explanation:

You just have to see how f(x) can be 0. In this expression f(x) = 0 when either of the parentheses is 0, i.e. either
(3x-5)=0
or (2x+7)=0
(3x-5)=0 when 3x=5 or $x = \frac{5}{3}$
(2x+7)=0 when 2x=-7 or $x = - \frac{7}{2} = - 3 \frac{1}{2}$