Solve 4x^2(x-3)-9(x-3)=0

$$4 { x }^{ 2 } (x-3)-9(x-3)=0$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=3$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Solution Steps

Use the distributive property to multiply 4x^{2} by x-3.

$$4x^{3}-12x^{2}-9\left(x-3\right)=0$$

Use the distributive property to multiply -9 by x-3.

$$4x^{3}-12x^{2}-9x+27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 27 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{27}{4},±\frac{27}{2},±27,±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}-12x^{2}-9x+27 by x-3 to get 4x^{2}-9. Solve the equation where the result equals to 0.

$$4x^{2}-9=0$$

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 4 for a, 0 for b, and -9 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 4\left(-9\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{0±12}{8}$$

Solve the equation 4x^{2}-9=0 when ± is plus and when ± is minus.

$$x=-\frac{3}{2} x=\frac{3}{2}$$

List all found solutions.

$$x=3 x=-\frac{3}{2} x=\frac{3}{2}$$