Solve a^2+10a-600

$$a ^ { 2 } + 10 a - 600$$

Factor

$$\left(a-20\right)\left(a+30\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as a^{2}+pa+qa-600. To find p and q, set up a system to be solved.

$$p+q=10 pq=1\left(-600\right)=-600$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -600.

$$-1,600 -2,300 -3,200 -4,150 -5,120 -6,100 -8,75 -10,60 -12,50 -15,40 -20,30 -24,25$$

Calculate the sum for each pair.

$$-1+600=599 -2+300=298 -3+200=197 -4+150=146 -5+120=115 -6+100=94 -8+75=67 -10+60=50 -12+50=38 -15+40=25 -20+30=10 -24+25=1$$

The solution is the pair that gives sum 10.

$$p=-20 q=30$$

Rewrite a^{2}+10a-600 as \left(a^{2}-20a\right)+\left(30a-600\right).

$$\left(a^{2}-20a\right)+\left(30a-600\right)$$

Factor out a in the first and 30 in the second group.

$$a\left(a-20\right)+30\left(a-20\right)$$

Factor out common term a-20 by using distributive property.

$$\left(a-20\right)\left(a+30\right)$$

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.

$$a^{2}+10a-600=0$$

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.

$$a=\frac{-10±\sqrt{10^{2}-4\left(-600\right)}}{2}$$

Square 10.

$$a=\frac{-10±\sqrt{100-4\left(-600\right)}}{2}$$

Multiply -4 times -600.

$$a=\frac{-10±\sqrt{100+2400}}{2}$$

Add 100 to 2400.

$$a=\frac{-10±\sqrt{2500}}{2}$$

Take the square root of 2500.

$$a=\frac{-10±50}{2}$$

Now solve the equation a=\frac{-10±50}{2} when ± is plus. Add -10 to 50.

$$a=\frac{40}{2}$$

Divide 40 by 2.

$$a=20$$

Now solve the equation a=\frac{-10±50}{2} when ± is minus. Subtract 50 from -10.

$$a=-\frac{60}{2}$$

Divide -60 by 2.

$$a=-30$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 20 for x_{1} and -30 for x_{2}.

$$a^{2}+10a-600=\left(a-20\right)\left(a-\left(-30\right)\right)$$

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

$$a^{2}+10a-600=\left(a-20\right)\left(a+30\right)$$

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.

$$x ^ 2 +10x -600 = 0$$

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

$$r + s = -10 rs = -600$$

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 = -600

$$(-5 - u) (-5 + u) = -600$$

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

$$25 - u^2 = -600$$

Simplify the expression by subtracting 25 on both sides

$$-u^2 = -600-25 = -625$$

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

$$u^2 = 625 u = \pm\sqrt{625} = \pm 25$$

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

$$r =-5 - 25 = -30 s = -5 + 25 = 20$$

Evaluate

$$\left(a-20\right)\left(a+30\right)$$