Solve x^2+10x+9=

$$x ^ { 2 } + 10 x + 9 =$$

Factor

$$\left(x+1\right)\left(x+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\times 9=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=1 b=9$$

Rewrite x^{2}+10x+9 as \left(x^{2}+x\right)+\left(9x+9\right).

$$\left(x^{2}+x\right)+\left(9x+9\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(x+1\right)+9\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+9\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}+10x+9=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-10±\sqrt{10^{2}-4\times 9}}{2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 9}}{2}$$

Multiply -4 times 9.

$$x=\frac{-10±\sqrt{100-36}}{2}$$

Add 100 to -36.

$$x=\frac{-10±\sqrt{64}}{2}$$

Take the square root of 64.

$$x=\frac{-10±8}{2}$$

Now solve the equation x=\frac{-10±8}{2} when ± is plus. Add -10 to 8.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-10±8}{2} when ± is minus. Subtract 8 from -10.

$$x=-\frac{18}{2}$$

Divide -18 by 2.

$$x=-9$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and -9 for x_{2}.

$$x^{2}+10x+9=\left(x-\left(-1\right)\right)\left(x-\left(-9\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$x^{2}+10x+9=\left(x+1\right)\left(x+9\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +10x +9 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -10 rs = 9$$

Two numbers r and s sum up to -10 exactly when the average of the two numbers is \frac{1}{2}*-10 = -5. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -5 - u s = -5 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 9

$$(-5 - u) (-5 + u) = 9$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$25 - u^2 = 9$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = 9-25 = -16$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = 16 u = \pm\sqrt{16} = \pm 4$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-5 - 4 = -9 s = -5 + 4 = -1$$

Evaluate

$$\left(x+1\right)\left(x+9\right)$$