Solve 3(x+1)^2+5(x+1)(x-4)+2(x-4)^2=0

$$3 ( x + 1 ) ^ { 2 } + 5 ( x + 1 ) ( x - 4 ) + 2 ( x - 4 ) ^ { 2 } = 0$$

Solve for x

$$x=1$$

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

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$3\left(x^{2}+2x+1\right)+5\left(x+1\right)\left(x-4\right)+2\left(x-4\right)^{2}=0$$

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

$$3x^{2}+6x+3+5\left(x+1\right)\left(x-4\right)+2\left(x-4\right)^{2}=0$$

Use the distributive property to multiply 5 by x+1.

$$3x^{2}+6x+3+\left(5x+5\right)\left(x-4\right)+2\left(x-4\right)^{2}=0$$

Use the distributive property to multiply 5x+5 by x-4 and combine like terms.

$$3x^{2}+6x+3+5x^{2}-15x-20+2\left(x-4\right)^{2}=0$$

Combine 3x^{2} and 5x^{2} to get 8x^{2}.

$$8x^{2}+6x+3-15x-20+2\left(x-4\right)^{2}=0$$

Combine 6x and -15x to get -9x.

$$8x^{2}-9x+3-20+2\left(x-4\right)^{2}=0$$

Subtract 20 from 3 to get -17.

$$8x^{2}-9x-17+2\left(x-4\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$8x^{2}-9x-17+2\left(x^{2}-8x+16\right)=0$$

Use the distributive property to multiply 2 by x^{2}-8x+16.

$$8x^{2}-9x-17+2x^{2}-16x+32=0$$

Combine 8x^{2} and 2x^{2} to get 10x^{2}.

$$10x^{2}-9x-17-16x+32=0$$

Combine -9x and -16x to get -25x.

$$10x^{2}-25x-17+32=0$$

Add -17 and 32 to get 15.

$$10x^{2}-25x+15=0$$

Divide both sides by 5.

$$2x^{2}-5x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-5 ab=2\times 3=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite 2x^{2}-5x+3 as \left(2x^{2}-3x\right)+\left(-2x+3\right).

$$\left(2x^{2}-3x\right)+\left(-2x+3\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(2x-3\right)-\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(x-1\right)$$

To find equation solutions, solve 2x-3=0 and x-1=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$3\left(x^{2}+2x+1\right)+5\left(x+1\right)\left(x-4\right)+2\left(x-4\right)^{2}=0$$

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

$$3x^{2}+6x+3+5\left(x+1\right)\left(x-4\right)+2\left(x-4\right)^{2}=0$$

Use the distributive property to multiply 5 by x+1.

$$3x^{2}+6x+3+\left(5x+5\right)\left(x-4\right)+2\left(x-4\right)^{2}=0$$

Use the distributive property to multiply 5x+5 by x-4 and combine like terms.

$$3x^{2}+6x+3+5x^{2}-15x-20+2\left(x-4\right)^{2}=0$$

Combine 3x^{2} and 5x^{2} to get 8x^{2}.

$$8x^{2}+6x+3-15x-20+2\left(x-4\right)^{2}=0$$

Combine 6x and -15x to get -9x.

$$8x^{2}-9x+3-20+2\left(x-4\right)^{2}=0$$

Subtract 20 from 3 to get -17.

$$8x^{2}-9x-17+2\left(x-4\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$8x^{2}-9x-17+2\left(x^{2}-8x+16\right)=0$$

Use the distributive property to multiply 2 by x^{2}-8x+16.

$$8x^{2}-9x-17+2x^{2}-16x+32=0$$

Combine 8x^{2} and 2x^{2} to get 10x^{2}.

$$10x^{2}-9x-17-16x+32=0$$

Combine -9x and -16x to get -25x.

$$10x^{2}-25x-17+32=0$$

Add -17 and 32 to get 15.

$$10x^{2}-25x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -25 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 10\times 15}}{2\times 10}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 10\times 15}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-\left(-25\right)±\sqrt{625-40\times 15}}{2\times 10}$$

Multiply -40 times 15.

$$x=\frac{-\left(-25\right)±\sqrt{625-600}}{2\times 10}$$

Add 625 to -600.

$$x=\frac{-\left(-25\right)±\sqrt{25}}{2\times 10}$$

Take the square root of 25.

$$x=\frac{-\left(-25\right)±5}{2\times 10}$$

The opposite of -25 is 25.

$$x=\frac{25±5}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{25±5}{20}$$

Now solve the equation x=\frac{25±5}{20} when ± is plus. Add 25 to 5.

$$x=\frac{30}{20}$$

Reduce the fraction \frac{30}{20} to lowest terms by extracting and canceling out 10.

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

Now solve the equation x=\frac{25±5}{20} when ± is minus. Subtract 5 from 25.

$$x=\frac{20}{20}$$

Divide 20 by 20.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$3\left(x^{2}+2x+1\right)+5\left(x+1\right)\left(x-4\right)+2\left(x-4\right)^{2}=0$$

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

$$3x^{2}+6x+3+5\left(x+1\right)\left(x-4\right)+2\left(x-4\right)^{2}=0$$

Use the distributive property to multiply 5 by x+1.

$$3x^{2}+6x+3+\left(5x+5\right)\left(x-4\right)+2\left(x-4\right)^{2}=0$$

Use the distributive property to multiply 5x+5 by x-4 and combine like terms.

$$3x^{2}+6x+3+5x^{2}-15x-20+2\left(x-4\right)^{2}=0$$

Combine 3x^{2} and 5x^{2} to get 8x^{2}.

$$8x^{2}+6x+3-15x-20+2\left(x-4\right)^{2}=0$$

Combine 6x and -15x to get -9x.

$$8x^{2}-9x+3-20+2\left(x-4\right)^{2}=0$$

Subtract 20 from 3 to get -17.

$$8x^{2}-9x-17+2\left(x-4\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$8x^{2}-9x-17+2\left(x^{2}-8x+16\right)=0$$

Use the distributive property to multiply 2 by x^{2}-8x+16.

$$8x^{2}-9x-17+2x^{2}-16x+32=0$$

Combine 8x^{2} and 2x^{2} to get 10x^{2}.

$$10x^{2}-9x-17-16x+32=0$$

Combine -9x and -16x to get -25x.

$$10x^{2}-25x-17+32=0$$

Add -17 and 32 to get 15.

$$10x^{2}-25x+15=0$$

Subtract 15 from both sides. Anything subtracted from zero gives its negation.

$$10x^{2}-25x=-15$$

Divide both sides by 10.

$$\frac{10x^{2}-25x}{10}=-\frac{15}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\left(-\frac{25}{10}\right)x=-\frac{15}{10}$$

Reduce the fraction \frac{-25}{10} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{5}{2}x=-\frac{15}{10}$$

Reduce the fraction \frac{-15}{10} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{5}{2}x=-\frac{3}{2}$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-\frac{3}{2}+\frac{25}{16}$$

Add -\frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{1}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{1}{4} x-\frac{5}{4}=-\frac{1}{4}$$

Add \frac{5}{4} to both sides of the equation.

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