Solve (2x^2+1)(2x+5)=(2x^2+1)(x-1)

$$( 2 x ^ { 2 } + 1 ) ( 2 x + 5 ) = ( 2 x ^ { 2 } + 1 ) ( x - 1 )$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{2}i}{2}\approx -0-0.707106781i$$

$$x=\frac{\sqrt{2}i}{2}\approx 0.707106781i$$

$$x=-6$$

Solution Steps

Use the distributive property to multiply 2x^{2}+1 by 2x+5.

$$4x^{3}+10x^{2}+2x+5=\left(2x^{2}+1\right)\left(x-1\right)$$

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

$$4x^{3}+10x^{2}+2x+5=2x^{3}-2x^{2}+x-1$$

Subtract 2x^{3} from both sides.

$$4x^{3}+10x^{2}+2x+5-2x^{3}=-2x^{2}+x-1$$

Combine 4x^{3} and -2x^{3} to get 2x^{3}.

$$2x^{3}+10x^{2}+2x+5=-2x^{2}+x-1$$

Add 2x^{2} to both sides.

$$2x^{3}+10x^{2}+2x+5+2x^{2}=x-1$$

Combine 10x^{2} and 2x^{2} to get 12x^{2}.

$$2x^{3}+12x^{2}+2x+5=x-1$$

Subtract x from both sides.

$$2x^{3}+12x^{2}+2x+5-x=-1$$

Combine 2x and -x to get x.

$$2x^{3}+12x^{2}+x+5=-1$$

Add 1 to both sides.

$$2x^{3}+12x^{2}+x+5+1=0$$

Add 5 and 1 to get 6.

$$2x^{3}+12x^{2}+x+6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

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

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.

$$x=-6$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+12x^{2}+x+6 by x+6 to get 2x^{2}+1. Solve the equation where the result equals to 0.

$$2x^{2}+1=0$$

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 2 for a, 0 for b, and 1 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\times 1}}{2\times 2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-8}}{4}$$

Solve the equation 2x^{2}+1=0 when ± is plus and when ± is minus.

$$x=-\frac{\sqrt{2}i}{2} x=\frac{\sqrt{2}i}{2}$$

List all found solutions.

$$x=-6 x=-\frac{\sqrt{2}i}{2} x=\frac{\sqrt{2}i}{2}$$

Solve for x

$$x=-6$$

Solution Steps

Use the distributive property to multiply 2x^{2}+1 by 2x+5.

$$4x^{3}+10x^{2}+2x+5=\left(2x^{2}+1\right)\left(x-1\right)$$

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

$$4x^{3}+10x^{2}+2x+5=2x^{3}-2x^{2}+x-1$$

Subtract 2x^{3} from both sides.

$$4x^{3}+10x^{2}+2x+5-2x^{3}=-2x^{2}+x-1$$

Combine 4x^{3} and -2x^{3} to get 2x^{3}.

$$2x^{3}+10x^{2}+2x+5=-2x^{2}+x-1$$

Add 2x^{2} to both sides.

$$2x^{3}+10x^{2}+2x+5+2x^{2}=x-1$$

Combine 10x^{2} and 2x^{2} to get 12x^{2}.

$$2x^{3}+12x^{2}+2x+5=x-1$$

Subtract x from both sides.

$$2x^{3}+12x^{2}+2x+5-x=-1$$

Combine 2x and -x to get x.

$$2x^{3}+12x^{2}+x+5=-1$$

Add 1 to both sides.

$$2x^{3}+12x^{2}+x+5+1=0$$

Add 5 and 1 to get 6.

$$2x^{3}+12x^{2}+x+6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

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

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.

$$x=-6$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+12x^{2}+x+6 by x+6 to get 2x^{2}+1. Solve the equation where the result equals to 0.

$$2x^{2}+1=0$$

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 2 for a, 0 for b, and 1 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\times 1}}{2\times 2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-8}}{4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-6$$