Solve a^2+11a+10=

$$a ^ { 2 } + 11 a + 10 =$$

Factor

$$\left(a+1\right)\left(a+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa+10. To find p and q, set up a system to be solved.

$$p+q=11 pq=1\times 10=10$$

Since pq is positive, p and q have the same sign. Since p+q is positive, p and q are both positive. List all such integer pairs that give product 10.

$$1,10 2,5$$

Calculate the sum for each pair.

$$1+10=11 2+5=7$$

The solution is the pair that gives sum 11.

$$p=1 q=10$$

Rewrite a^{2}+11a+10 as \left(a^{2}+a\right)+\left(10a+10\right).

$$\left(a^{2}+a\right)+\left(10a+10\right)$$

Factor out a in the first and 10 in the second group.

$$a\left(a+1\right)+10\left(a+1\right)$$

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

$$\left(a+1\right)\left(a+10\right)$$

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.

$$a^{2}+11a+10=0$$

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.

$$a=\frac{-11±\sqrt{11^{2}-4\times 10}}{2}$$

Square 11.

$$a=\frac{-11±\sqrt{121-4\times 10}}{2}$$

Multiply -4 times 10.

$$a=\frac{-11±\sqrt{121-40}}{2}$$

Add 121 to -40.

$$a=\frac{-11±\sqrt{81}}{2}$$

Take the square root of 81.

$$a=\frac{-11±9}{2}$$

Now solve the equation a=\frac{-11±9}{2} when ± is plus. Add -11 to 9.

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

Divide -2 by 2.

$$a=-1$$

Now solve the equation a=\frac{-11±9}{2} when ± is minus. Subtract 9 from -11.

$$a=-\frac{20}{2}$$

Divide -20 by 2.

$$a=-10$$

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 -10 for x_{2}.

$$a^{2}+11a+10=\left(a-\left(-1\right)\right)\left(a-\left(-10\right)\right)$$

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

$$a^{2}+11a+10=\left(a+1\right)\left(a+10\right)$$

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.

$$x ^ 2 +11x +10 = 0$$

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

$$r + s = -11 rs = 10$$

Two numbers r and s sum up to -11 exactly when the average of the two numbers is \frac{1}{2}*-11 = -\frac{11}{2}. 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>

$$r = -\frac{11}{2} - u s = -\frac{11}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 10

$$(-\frac{11}{2} - u) (-\frac{11}{2} + u) = 10$$

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

$$\frac{121}{4} - u^2 = 10$$

Simplify the expression by subtracting \frac{121}{4} on both sides

$$-u^2 = 10-\frac{121}{4} = -\frac{81}{4}$$

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

$$u^2 = \frac{81}{4} u = \pm\sqrt{\frac{81}{4}} = \pm \frac{9}{2}$$

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

$$r =-\frac{11}{2} - \frac{9}{2} = -10 s = -\frac{11}{2} + \frac{9}{2} = -1$$

Evaluate

$$\left(a+1\right)\left(a+10\right)$$