Question

factors and zeros
find all zeros.

1. ( f(x)=(2 x - 1)(x - 5) )
2. ( f(x)=(x - 3)(3 x + 1)(x + 1) )
3. ( f(x)=(2 x + 1)(x + 1)(x - 1) )
4. ( f(x)=x(5 x - 2)(x^{2}+1) )
5. ( f(x)=x(x + 2)(x - 2)(3 x^{2}-4) )
6. ( f(x)=(2 x - 1)(x^{2}+3)(2 x^{2}-5) )
7. ( f(x)=x(2 x - 1)(x - 1)(x + 1) )
8. ( f(x)=(2 x + 5)(x^{2}-2 x - 5) )

write a polynomial function of least degree with integral coefficients that has the given zeros.

1. ( 3,2,-2 )
2. ( 3,1,-2,-4 )

Explanation:

Step1: Use the zero - product property

If \(f(x)=(2x - 1)(x - 5)\), then set each factor equal to zero.
For \(2x-1 = 0\), we solve for \(x\):
\(2x-1=0\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)
For \(x - 5=0\), we solve for \(x\):
\(x-5=0\Rightarrow x = 5\)

Step2: Check the solutions

Substitute \(x=\frac{1}{2}\) into \(f(x)\):
\(f(\frac{1}{2})=(2\times\frac{1}{2}-1)(\frac{1}{2}-5)=(1 - 1)(\frac{1}{2}-5)=0\times(-\frac{9}{2})=0\)
Substitute \(x = 5\) into \(f(x)\):
\(f(5)=(2\times5-1)(5 - 5)=(10 - 1)\times0=0\)

Answer:

The zeros of \(f(x)=(2x - 1)(x - 5)\) are \(x=\frac{1}{2}\) and \(x = 5\)