Solve (x-5)(x-3)=0quadb)

$$( x - 5 ) ( x - 3 ) = 0 \quad b )$$

Solve for x

$$x=3$$

$$x=5$$

Steps Using Factoring

Use the distributive property to multiply x-5 by x-3 and combine like terms.

$$x^{2}-8x+15=0b$$

Anything times zero gives zero.

$$x^{2}-8x+15=0$$

To solve the equation, factor x^{2}-8x+15 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-8 ab=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

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

$$\left(x-5\right)\left(x-3\right)$$

To find equation solutions, solve x-5=0 and x-3=0.

$$x=5 x=3$$

Steps Using Factoring By Grouping

Use the distributive property to multiply x-5 by x-3 and combine like terms.

$$x^{2}-8x+15=0b$$

Anything times zero gives zero.

$$x^{2}-8x+15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+15. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 15=15$$

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 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -8.

$$a=-5 b=-3$$

Rewrite x^{2}-8x+15 as \left(x^{2}-5x\right)+\left(-3x+15\right).

$$\left(x^{2}-5x\right)+\left(-3x+15\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-5\right)-3\left(x-5\right)$$

Factor out common term x-5 by using distributive property.

$$\left(x-5\right)\left(x-3\right)$$

To find equation solutions, solve x-5=0 and x-3=0.

$$x=5 x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-5 by x-3 and combine like terms.

$$x^{2}-8x+15=0b$$

Anything times zero gives zero.

$$x^{2}-8x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 15}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 15}}{2}$$

Multiply -4 times 15.

$$x=\frac{-\left(-8\right)±\sqrt{64-60}}{2}$$

Add 64 to -60.

$$x=\frac{-\left(-8\right)±\sqrt{4}}{2}$$

Take the square root of 4.

$$x=\frac{-\left(-8\right)±2}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±2}{2}$$

Now solve the equation x=\frac{8±2}{2} when ± is plus. Add 8 to 2.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{8±2}{2} when ± is minus. Subtract 2 from 8.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

The equation is now solved.

$$x=5 x=3$$

Steps for Completing the Square

Use the distributive property to multiply x-5 by x-3 and combine like terms.

$$x^{2}-8x+15=0b$$

Anything times zero gives zero.

$$x^{2}-8x+15=0$$

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

$$x^{2}-8x=-15$$

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.

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

Square -4.

$$x^{2}-8x+16=-15+16$$

Add -15 to 16.

$$x^{2}-8x+16=1$$

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

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=1 x-4=-1$$

Add 4 to both sides of the equation.

$$x=5 x=3$$

Solve for b (complex solution)

$$b\in \mathrm{C}$$

$$x=3\text{ or }x=5$$

Solve for b

$$b\in \mathrm{R}$$

$$x=3\text{ or }x=5$$