Chapter 2, Problem 53CR

To determine

To find: rational zeros of function using rational zero test.

Answer to Problem 53CR

  $0,0,-2,-3$ .

Explanation of Solution

Given information:

  $f(x)=x^4+5x^3+6x^2=x^2(x^2+5x+6)$

  $=g(x)h(x)$

Concept used: - Rational root test:- Let f(x) be a polynomial and c be constant term in f(x) and a is leading coefficient of f(x) then ratio of possible divisors of constant term and leading coefficient can rational zeros of function.

Rational root = $\frac{divisors\text{ of constant term }}{divisors\text{ of leading cofficient}}$

1^stmethod:-

Since, we know zeros of a polynomial satisfies polynomial so, by putting these numbers in polynomial we can check which number is zero of polynomial.

2^ndmethod: - using synthetic division method we can also check the roots of polynomial.

Calculation: applying rational root test for h(x), we have

  $h(x)=(x^2+5x+6)$

Possible divisors of constant term c = 6 are $\pm 1,\pm 2,\pm 3,\pm 6$

And possible divisors of leading coefficient term is 1

  $\begin{array}{l}h(x)=(x^2+5x+6)\\ h(1)=1^2+5+6=12\ne 0\\ $ h(-1)={(-1)}^2+5(-1)+6=1-5+6=2\ne 0$h(2)=2^2+5\times 2+6=4+10+6\\ =20\ne 0\\ $ h(-2)={(-2)}^2+5\times (-2)+6=4-10+6=0$h(3)=3^2+5\times 3+6=9+15+6=30\ne 0\\ $ h(-3)={(-3)}^2+5\times (-3)+6=9-15+6=0$\end{array}$

Since, h(x) is 2 degree polynomial so it has 2 zeros -2, -3

(As we have found 2 zeros for 2 degree polynomial so no need to test further for zeros).

  $g(x)=x^2$

Zeros of g(x) = 0, 0

As $f(x)$ = $g(x)h(x)$

Therefore, zeros of f(x) is 0, 0, -2, -3.