Solve 5w^2+35w+50

$$5 w ^ { 2 } + 35 w + 50$$

Factor

$$5\left(w+2\right)\left(w+5\right)$$

Solution Steps

Factor out 5.

$$5\left(w^{2}+7w+10\right)$$

Consider w^{2}+7w+10. Factor the expression by grouping. First, the expression needs to be rewritten as w^{2}+aw+bw+10. To find a and b, set up a system to be solved.

$$a+b=7 ab=1\times 10=10$$

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 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 7.

$$a=2 b=5$$

Rewrite w^{2}+7w+10 as \left(w^{2}+2w\right)+\left(5w+10\right).

$$\left(w^{2}+2w\right)+\left(5w+10\right)$$

Factor out w in the first and 5 in the second group.

$$w\left(w+2\right)+5\left(w+2\right)$$

Factor out common term w+2 by using distributive property.

$$\left(w+2\right)\left(w+5\right)$$

Rewrite the complete factored expression.

$$5\left(w+2\right)\left(w+5\right)$$

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.

$$5w^{2}+35w+50=0$$

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.

$$w=\frac{-35±\sqrt{35^{2}-4\times 5\times 50}}{2\times 5}$$

Square 35.

$$w=\frac{-35±\sqrt{1225-4\times 5\times 50}}{2\times 5}$$

Multiply -4 times 5.

$$w=\frac{-35±\sqrt{1225-20\times 50}}{2\times 5}$$

Multiply -20 times 50.

$$w=\frac{-35±\sqrt{1225-1000}}{2\times 5}$$

Add 1225 to -1000.

$$w=\frac{-35±\sqrt{225}}{2\times 5}$$

Take the square root of 225.

$$w=\frac{-35±15}{2\times 5}$$

Multiply 2 times 5.

$$w=\frac{-35±15}{10}$$

Now solve the equation w=\frac{-35±15}{10} when ± is plus. Add -35 to 15.

$$w=-\frac{20}{10}$$

Divide -20 by 10.

$$w=-2$$

Now solve the equation w=\frac{-35±15}{10} when ± is minus. Subtract 15 from -35.

$$w=-\frac{50}{10}$$

Divide -50 by 10.

$$w=-5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -2 for x_{1} and -5 for x_{2}.

$$5w^{2}+35w+50=5\left(w-\left(-2\right)\right)\left(w-\left(-5\right)\right)$$

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

$$5w^{2}+35w+50=5\left(w+2\right)\left(w+5\right)$$

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.This is achieved by dividing both sides of the equation by 5

$$x ^ 2 +7x +10 = 0$$

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

$$r + s = -7 rs = 10$$

Two numbers r and s sum up to -7 exactly when the average of the two numbers is \frac{1}{2}*-7 = -\frac{7}{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>

$$r = -\frac{7}{2} - u s = -\frac{7}{2} + u$$

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

$$(-\frac{7}{2} - u) (-\frac{7}{2} + u) = 10$$

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

$$\frac{49}{4} - u^2 = 10$$

Simplify the expression by subtracting \frac{49}{4} on both sides

$$-u^2 = 10-\frac{49}{4} = -\frac{9}{4}$$

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

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

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

$$r =-\frac{7}{2} - \frac{3}{2} = -5 s = -\frac{7}{2} + \frac{3}{2} = -2$$

Evaluate

$$5\left(w+2\right)\left(w+5\right)$$