### If(x-4) is a factor of a quadratic polynomial p(x) and 2 is zero of p(x), then find the polynomial p(x).

**(a) \(x^2+6x+8\)**

**(b) \(x^2-6x+8\)**

**(c) \(x^2+x+8\)**

**(d) \(x^2-x+8\)**

**Answer: Option (b) \(x^2-6x+8\)**

**Solution:**

**Step 1: Use the given information about the roots of the polynomial.**

**By factor theorem:** We know if \(x-a\) is factor of a polynomial \(p(x)\) then \(p(a)=0\), that is \(x=a\) is zero/root of \(p(x)\).

We have given that \((x - 4)\) is a factor of the quadratic polynomial \(p(x)\).

So, \(x = 4\) is zero/root of \(p(x)\).

Also, given that \(x = 2\) is a zero/root of \(p(x)\).

So, the roots of the polynomial are \(x = 4\) and \(x = 2\).

**Step 2: Use the factor form of a quadratic polynomial.**

The quadratic polynomial whose roots are \(x = 4\) and \(x = 2\) is given by:

\(

p(x) = k(x - 4)(x - 2)

\)

Here, \(kÂ = 1\) because we are assuming the polynomial is monic (i.e., leading coefficient is 1). Therefore, the polynomial becomes:

\(

p(x) = (x - 4)(x - 2)

\)

**Step 3: Expand the factored form.**

Now, expand \((x - 4)(x - 2)\):

\(

p(x) = (x - 4)(x - 2) = x^2 - 2x - 4x + 8Â \newline \Rightarrow p(x)= x^2 - 6x + 8

\)

The quadratic polynomial is:

\(

p(x) = x^2 - 6x + 8

\)

**Hence, the correct answer is option (b) \(x^2 - 6x + 8\).**