Given:
The polynomial is
[tex]p(x)=x^2+11x-60[/tex]
To find:
Whether x = 4 is a zero of the given polynomial by remainder theorem and find the quotient and remainder.
Solution:
According to the remainder theorem, if (x-c) divides a polynomial p(x), then the remainder is p(c).
Divide the polynomial by (x-4) and check whether the remainder p(x)=0 at x=4.
Putting x=4 in the given polynomial, we get
[tex]p(4)=(4)^2+11(4)-60[/tex]
[tex]p(4)=16+44-60[/tex]
[tex]p(4)=60-60[/tex]
[tex]p(4)=0[/tex]
Since, remainder is 0, therefore, (x-4) is a factor of p(x) and x=4 is a zero.
Now, on dividing p(x) be (x-4) we get
[tex]\dfrac{x^2+11x-60}{x-4}=\dfrac{x^2+15x-4x-60}{x-4}[/tex]
[tex]\dfrac{x^2+11x-60}{x-4}=\dfrac{x(x+15)-4(x+15)}{x-4}[/tex]
[tex]\dfrac{x^2+11x-60}{x-4}=\dfrac{(x+15)(x-4)}{x-4}[/tex]
[tex]\dfrac{x^2+11x-60}{x-4}=x+15[/tex]
The quotient is x+15 and remainder is 0.
Therefore, the correct option is A.
