Solve (x-3)(x-25)(x+1)=03

$$(x-3)(x-25)(x+1)=03$$

Solve for x

$$x=25$$

$$x=3$$

$$x=-1$$

Solution Steps

Use the distributive property to multiply x-3 by x-25 and combine like terms.

$$\left(x^{2}-28x+75\right)\left(x+1\right)=0\times 3$$

Use the distributive property to multiply x^{2}-28x+75 by x+1 and combine like terms.

$$x^{3}-27x^{2}+47x+75=0\times 3$$

Multiply 0 and 3 to get 0.

$$x^{3}-27x^{2}+47x+75=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 75 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±75,±25,±15,±5,±3,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-27x^{2}+47x+75 by x+1 to get x^{2}-28x+75. Solve the equation where the result equals to 0.

$$x^{2}-28x+75=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -28 for b, and 75 for c in the quadratic formula.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 1\times 75}}{2}$$

Do the calculations.

$$x=\frac{28±22}{2}$$

Solve the equation x^{2}-28x+75=0 when ± is plus and when ± is minus.

$$x=3 x=25$$

List all found solutions.

$$x=-1 x=3 x=25$$