Solve f(z)=18z^3-9z^2-50z+25

$$f ( z ) = 18 z ^ { 3 } - 9 z ^ { 2 } - 50 z + 25$$

Factor

$$\left(2z-1\right)\left(3z-5\right)\left(3z+5\right)$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 25 and q divides the leading coefficient 18. One such root is \frac{5}{3}. Factor the polynomial by dividing it by 3z-5.

$$\left(3z-5\right)\left(6z^{2}+7z-5\right)$$

Consider 6z^{2}+7z-5. Factor the expression by grouping. First, the expression needs to be rewritten as 6z^{2}+az+bz-5. To find a and b, set up a system to be solved.

$$a+b=7 ab=6\left(-5\right)=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 7.

$$a=-3 b=10$$

Rewrite 6z^{2}+7z-5 as \left(6z^{2}-3z\right)+\left(10z-5\right).

$$\left(6z^{2}-3z\right)+\left(10z-5\right)$$

Factor out 3z in the first and 5 in the second group.

$$3z\left(2z-1\right)+5\left(2z-1\right)$$

Factor out common term 2z-1 by using distributive property.

$$\left(2z-1\right)\left(3z+5\right)$$

Rewrite the complete factored expression.

$$\left(3z-5\right)\left(2z-1\right)\left(3z+5\right)$$

Evaluate

$$\left(2z-1\right)\left(9z^{2}-25\right)$$