Question

Let ​ f(x)=x^2+17x+72 What are the zeros of the function?

Answer 1

Answer:

The zeros of the given function  $f(x)=x^2+17x+72$ is $\left(x+8\right)\left(x+9\right)$ are -8 and -9.

Step-by-step explanation:

Given : Function  $f(x)=x^2+17x+72$

We have to find the zeros of the functions.

Consider the given  Function  $f(x)=x^2+17x+72$

Since, we have to find the zeros of the given functions.

Put f(x) = 0

Thus,  $x^2+17x+72=0$

We can solve the given equation using middle term splitting method.

17x can be written as 8x + 9x

$x^2+17x+72=0$ can be written as $x^2+8x+9x+72=0$

Taking x common from first two term and 9 common from last two terms, we have,

$x(x+8)+9(x+8)=0$

Simplify, we have,

$\left(x+8\right)=0$ or $(x+9\right)=0$

x = -8 and x = -9

Apply zero product rule,$a\cdot b= 0 \Rightarrow a=0 \ or \ b=0$

$\left(x+8\right)\left(x+9\right)=0$

Thus, The zero of the given function  $f(x)=x^2+17x+72$ is $\left(x+8\right)\left(x+9\right)$ are -8 and -9.