Solve (3x-1)(x^2+2)=(3x-1)(7x-10)

$$( 3 x - 1 ) ( x ^ { 2 } + 2 ) = ( 3 x - 1 ) ( 7 x - 10 )$$

Solve for x

$$x=3$$

$$x=4$$

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

Solution Steps

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

$$3x^{3}+6x-x^{2}-2=\left(3x-1\right)\left(7x-10\right)$$

Use the distributive property to multiply 3x-1 by 7x-10 and combine like terms.

$$3x^{3}+6x-x^{2}-2=21x^{2}-37x+10$$

Subtract 21x^{2} from both sides.

$$3x^{3}+6x-x^{2}-2-21x^{2}=-37x+10$$

Combine -x^{2} and -21x^{2} to get -22x^{2}.

$$3x^{3}+6x-22x^{2}-2=-37x+10$$

Add 37x to both sides.

$$3x^{3}+6x-22x^{2}-2+37x=10$$

Combine 6x and 37x to get 43x.

$$3x^{3}+43x-22x^{2}-2=10$$

Subtract 10 from both sides.

$$3x^{3}+43x-22x^{2}-2-10=0$$

Subtract 10 from -2 to get -12.

$$3x^{3}+43x-22x^{2}-12=0$$

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

$$3x^{3}-22x^{2}+43x-12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -12 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±4,±12,±2,±6,±\frac{4}{3},±1,±3,±\frac{2}{3},±\frac{1}{3}$$

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 3x^{3}-22x^{2}+43x-12 by x-3 to get 3x^{2}-13x+4. Solve the equation where the result equals to 0.

$$3x^{2}-13x+4=0$$

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, -13 for b, and 4 for c in the quadratic formula.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 3\times 4}}{2\times 3}$$

Do the calculations.

$$x=\frac{13±11}{6}$$

Solve the equation 3x^{2}-13x+4=0 when ± is plus and when ± is minus.

$$x=\frac{1}{3} x=4$$

List all found solutions.

$$x=3 x=\frac{1}{3} x=4$$