Question

Please help!

Answer 1

Answer:

The zeros are:

$x =4, x=2, x = 5$

• The function has three distinct real zeros.

Hence, option (B) is true.

Step-by-step explanation:

Given the expression

$h\left(x\right)=\left(x-4\right)^2\left(x^2-7x+\:10\right)$

Let us determine the zeros of the function by putting h(x) = 0 and solving the expression

$0=\left(x-4\right)^2\left(x^2-7x+10\right)$

switch sides

$\left(x-4\right)^2\left(x^2-7x+10\right)=0$

as

$x^2-7x+\:10=\left(x-2\right)\left(x-5\right)$

so

$\left(x-4\right)^2\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

• $\mathrm{If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

so

$x-4=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x =4, x=2, x = 5$

Thus, the zeros are:

$x =4, x=2, x = 5$

It is clear that there are three zeros and all the zeros are distinct real numbers.

Therefore,

• The function has three distinct real zeros.

Hence, option (B) is true.