Solve (m^2-10)*(m-4)=(m^2-2m-8)(m-3)

$$( m ^ { 2 } - 10 ) \cdot ( m - 4 ) = ( m ^ { 2 } - 2 m - 8 ) ( m - 3 )$$

Solve for m

$$m=4$$

Steps Using Factoring

Use the distributive property to multiply m^{2}-10 by m-4.

$$m^{3}-4m^{2}-10m+40=\left(m^{2}-2m-8\right)\left(m-3\right)$$

Use the distributive property to multiply m^{2}-2m-8 by m-3 and combine like terms.

$$m^{3}-4m^{2}-10m+40=m^{3}-5m^{2}-2m+24$$

Subtract m^{3} from both sides.

$$m^{3}-4m^{2}-10m+40-m^{3}=-5m^{2}-2m+24$$

Combine m^{3} and -m^{3} to get 0.

$$-4m^{2}-10m+40=-5m^{2}-2m+24$$

Add 5m^{2} to both sides.

$$-4m^{2}-10m+40+5m^{2}=-2m+24$$

Combine -4m^{2} and 5m^{2} to get m^{2}.

$$m^{2}-10m+40=-2m+24$$

Add 2m to both sides.

$$m^{2}-10m+40+2m=24$$

Combine -10m and 2m to get -8m.

$$m^{2}-8m+40=24$$

Subtract 24 from both sides.

$$m^{2}-8m+40-24=0$$

Subtract 24 from 40 to get 16.

$$m^{2}-8m+16=0$$

To solve the equation, factor m^{2}-8m+16 using formula m^{2}+\left(a+b\right)m+ab=\left(m+a\right)\left(m+b\right). To find a and b, set up a system to be solved.

$$a+b=-8 ab=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$a=-4 b=-4$$

Rewrite factored expression \left(m+a\right)\left(m+b\right) using the obtained values.

$$\left(m-4\right)\left(m-4\right)$$

Rewrite as a binomial square.

$$\left(m-4\right)^{2}$$

To find equation solution, solve m-4=0.

$$m=4$$

Steps Using Factoring By Grouping

Use the distributive property to multiply m^{2}-10 by m-4.

$$m^{3}-4m^{2}-10m+40=\left(m^{2}-2m-8\right)\left(m-3\right)$$

Use the distributive property to multiply m^{2}-2m-8 by m-3 and combine like terms.

$$m^{3}-4m^{2}-10m+40=m^{3}-5m^{2}-2m+24$$

Subtract m^{3} from both sides.

$$m^{3}-4m^{2}-10m+40-m^{3}=-5m^{2}-2m+24$$

Combine m^{3} and -m^{3} to get 0.

$$-4m^{2}-10m+40=-5m^{2}-2m+24$$

Add 5m^{2} to both sides.

$$-4m^{2}-10m+40+5m^{2}=-2m+24$$

Combine -4m^{2} and 5m^{2} to get m^{2}.

$$m^{2}-10m+40=-2m+24$$

Add 2m to both sides.

$$m^{2}-10m+40+2m=24$$

Combine -10m and 2m to get -8m.

$$m^{2}-8m+40=24$$

Subtract 24 from both sides.

$$m^{2}-8m+40-24=0$$

Subtract 24 from 40 to get 16.

$$m^{2}-8m+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as m^{2}+am+bm+16. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 16=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Calculate the sum for each pair.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

The solution is the pair that gives sum -8.

$$a=-4 b=-4$$

Rewrite m^{2}-8m+16 as \left(m^{2}-4m\right)+\left(-4m+16\right).

$$\left(m^{2}-4m\right)+\left(-4m+16\right)$$

Factor out m in the first and -4 in the second group.

$$m\left(m-4\right)-4\left(m-4\right)$$

Factor out common term m-4 by using distributive property.

$$\left(m-4\right)\left(m-4\right)$$

Rewrite as a binomial square.

$$\left(m-4\right)^{2}$$

To find equation solution, solve m-4=0.

$$m=4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply m^{2}-10 by m-4.

$$m^{3}-4m^{2}-10m+40=\left(m^{2}-2m-8\right)\left(m-3\right)$$

Use the distributive property to multiply m^{2}-2m-8 by m-3 and combine like terms.

$$m^{3}-4m^{2}-10m+40=m^{3}-5m^{2}-2m+24$$

Subtract m^{3} from both sides.

$$m^{3}-4m^{2}-10m+40-m^{3}=-5m^{2}-2m+24$$

Combine m^{3} and -m^{3} to get 0.

$$-4m^{2}-10m+40=-5m^{2}-2m+24$$

Add 5m^{2} to both sides.

$$-4m^{2}-10m+40+5m^{2}=-2m+24$$

Combine -4m^{2} and 5m^{2} to get m^{2}.

$$m^{2}-10m+40=-2m+24$$

Add 2m to both sides.

$$m^{2}-10m+40+2m=24$$

Combine -10m and 2m to get -8m.

$$m^{2}-8m+40=24$$

Subtract 24 from both sides.

$$m^{2}-8m+40-24=0$$

Subtract 24 from 40 to get 16.

$$m^{2}-8m+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 16}}{2}$$

Square -8.

$$m=\frac{-\left(-8\right)±\sqrt{64-4\times 16}}{2}$$

Multiply -4 times 16.

$$m=\frac{-\left(-8\right)±\sqrt{64-64}}{2}$$

Add 64 to -64.

$$m=\frac{-\left(-8\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$m=-\frac{-8}{2}$$

The opposite of -8 is 8.

$$m=\frac{8}{2}$$

Divide 8 by 2.

$$m=4$$

Steps for Completing the Square

Use the distributive property to multiply m^{2}-10 by m-4.

$$m^{3}-4m^{2}-10m+40=\left(m^{2}-2m-8\right)\left(m-3\right)$$

Use the distributive property to multiply m^{2}-2m-8 by m-3 and combine like terms.

$$m^{3}-4m^{2}-10m+40=m^{3}-5m^{2}-2m+24$$

Subtract m^{3} from both sides.

$$m^{3}-4m^{2}-10m+40-m^{3}=-5m^{2}-2m+24$$

Combine m^{3} and -m^{3} to get 0.

$$-4m^{2}-10m+40=-5m^{2}-2m+24$$

Add 5m^{2} to both sides.

$$-4m^{2}-10m+40+5m^{2}=-2m+24$$

Combine -4m^{2} and 5m^{2} to get m^{2}.

$$m^{2}-10m+40=-2m+24$$

Add 2m to both sides.

$$m^{2}-10m+40+2m=24$$

Combine -10m and 2m to get -8m.

$$m^{2}-8m+40=24$$

Subtract 40 from both sides.

$$m^{2}-8m=24-40$$

Subtract 40 from 24 to get -16.

$$m^{2}-8m=-16$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}-8m+\left(-4\right)^{2}=-16+\left(-4\right)^{2}$$

Square -4.

$$m^{2}-8m+16=-16+16$$

Add -16 to 16.

$$m^{2}-8m+16=0$$

Factor m^{2}-8m+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}.

$$\left(m-4\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-4\right)^{2}}=\sqrt{0}$$

Simplify.

$$m-4=0 m-4=0$$

Add 4 to both sides of the equation.

$$m=4 m=4$$

The equation is now solved. Solutions are the same.

$$m=4$$