Solve (3x-1)(x^2+4)=(3x-1)(8x-3)

$$( 3 x - 1 ) ( x ^ { 2 } + 4 ) = ( 3 x - 1 ) ( 8 x - 3 )$$

Solve for x

$$x=1$$

$$x=7$$

$$x=\frac{1}{3}\approx 0.333333333$$

Solution Steps

Use the distributive property to multiply 3x-1 by x^{2}+4.

$$3x^{3}+12x-x^{2}-4=\left(3x-1\right)\left(8x-3\right)$$

Use the distributive property to multiply 3x-1 by 8x-3 and combine like terms.

$$3x^{3}+12x-x^{2}-4=24x^{2}-17x+3$$

Subtract 24x^{2} from both sides.

$$3x^{3}+12x-x^{2}-4-24x^{2}=-17x+3$$

Combine -x^{2} and -24x^{2} to get -25x^{2}.

$$3x^{3}+12x-25x^{2}-4=-17x+3$$

Add 17x to both sides.

$$3x^{3}+12x-25x^{2}-4+17x=3$$

Combine 12x and 17x to get 29x.

$$3x^{3}+29x-25x^{2}-4=3$$

Subtract 3 from both sides.

$$3x^{3}+29x-25x^{2}-4-3=0$$

Subtract 3 from -4 to get -7.

$$3x^{3}+29x-25x^{2}-7=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$3x^{3}-25x^{2}+29x-7=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -7 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{7}{3},±7,±\frac{1}{3},±1$$

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 3x^{3}-25x^{2}+29x-7 by x-1 to get 3x^{2}-22x+7. Solve the equation where the result equals to 0.

$$3x^{2}-22x+7=0$$

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 3 for a, -22 for b, and 7 for c in the quadratic formula.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 3\times 7}}{2\times 3}$$

Do the calculations.

$$x=\frac{22±20}{6}$$

Solve the equation 3x^{2}-22x+7=0 when ± is plus and when ± is minus.

$$x=\frac{1}{3} x=7$$

List all found solutions.

$$x=1 x=\frac{1}{3} x=7$$