Solve 5w^2+3w-14

$$5 w ^ { 2 } + 3 w - 14$$

Factor

$$\left(5w-7\right)\left(w+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5w^{2}+aw+bw-14. To find a and b, set up a system to be solved.

$$a+b=3 ab=5\left(-14\right)=-70$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -70.

$$-1,70 -2,35 -5,14 -7,10$$

Calculate the sum for each pair.

$$-1+70=69 -2+35=33 -5+14=9 -7+10=3$$

The solution is the pair that gives sum 3.

$$a=-7 b=10$$

Rewrite 5w^{2}+3w-14 as \left(5w^{2}-7w\right)+\left(10w-14\right).

$$\left(5w^{2}-7w\right)+\left(10w-14\right)$$

Factor out w in the first and 2 in the second group.

$$w\left(5w-7\right)+2\left(5w-7\right)$$

Factor out common term 5w-7 by using distributive property.

$$\left(5w-7\right)\left(w+2\right)$$

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.

$$5w^{2}+3w-14=0$$

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.

$$w=\frac{-3±\sqrt{3^{2}-4\times 5\left(-14\right)}}{2\times 5}$$

Square 3.

$$w=\frac{-3±\sqrt{9-4\times 5\left(-14\right)}}{2\times 5}$$

Multiply -4 times 5.

$$w=\frac{-3±\sqrt{9-20\left(-14\right)}}{2\times 5}$$

Multiply -20 times -14.

$$w=\frac{-3±\sqrt{9+280}}{2\times 5}$$

Add 9 to 280.

$$w=\frac{-3±\sqrt{289}}{2\times 5}$$

Take the square root of 289.

$$w=\frac{-3±17}{2\times 5}$$

Multiply 2 times 5.

$$w=\frac{-3±17}{10}$$

Now solve the equation w=\frac{-3±17}{10} when ± is plus. Add -3 to 17.

$$w=\frac{14}{10}$$

Reduce the fraction \frac{14}{10} to lowest terms by extracting and canceling out 2.

$$w=\frac{7}{5}$$

Now solve the equation w=\frac{-3±17}{10} when ± is minus. Subtract 17 from -3.

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

Divide -20 by 10.

$$w=-2$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{7}{5} for x_{1} and -2 for x_{2}.

$$5w^{2}+3w-14=5\left(w-\frac{7}{5}\right)\left(w-\left(-2\right)\right)$$

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

$$5w^{2}+3w-14=5\left(w-\frac{7}{5}\right)\left(w+2\right)$$

Subtract \frac{7}{5} from w by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5w^{2}+3w-14=5\times \frac{5w-7}{5}\left(w+2\right)$$

Cancel out 5, the greatest common factor in 5 and 5.

$$5w^{2}+3w-14=\left(5w-7\right)\left(w+2\right)$$

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

$$x ^ 2 +\frac{3}{5}x -\frac{14}{5} = 0$$

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

$$r + s = -\frac{3}{5} rs = -\frac{14}{5}$$

Two numbers r and s sum up to -\frac{3}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{5} = -\frac{3}{10}. 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>

$$r = -\frac{3}{10} - u s = -\frac{3}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{14}{5}

$$(-\frac{3}{10} - u) (-\frac{3}{10} + u) = -\frac{14}{5}$$

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

$$\frac{9}{100} - u^2 = -\frac{14}{5}$$

Simplify the expression by subtracting \frac{9}{100} on both sides

$$-u^2 = -\frac{14}{5}-\frac{9}{100} = -\frac{289}{100}$$

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

$$u^2 = \frac{289}{100} u = \pm\sqrt{\frac{289}{100}} = \pm \frac{17}{10}$$

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

$$r =-\frac{3}{10} - \frac{17}{10} = -2 s = -\frac{3}{10} + \frac{17}{10} = 1.400$$

Evaluate

$$\left(5w-7\right)\left(w+2\right)$$