Answer:
The zeros are:
[tex]x =4, x=2, x = 5[/tex]
- The function has three distinct real zeros.
Hence, option (B) is true.
Step-by-step explanation:
Given the expression
[tex]h\left(x\right)=\left(x-4\right)^2\left(x^2-7x+\:10\right)[/tex]
Let us determine the zeros of the function by putting h(x) = 0 and solving the expression
[tex]0=\left(x-4\right)^2\left(x^2-7x+10\right)[/tex]
switch sides
[tex]\left(x-4\right)^2\left(x^2-7x+10\right)=0[/tex]
as
[tex]x^2-7x+\:10=\left(x-2\right)\left(x-5\right)[/tex]
so
[tex]\left(x-4\right)^2\left(x-2\right)\left(x-5\right)=0[/tex]
Using the zero factor principle
- [tex]\mathrm{If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)[/tex]
so
[tex]x-4=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0[/tex]
[tex]x =4, x=2, x = 5[/tex]
Thus, the zeros are:
[tex]x =4, x=2, x = 5[/tex]
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.