Solve -18a^2-102a+168

$$- 18 a ^ { 2 } - 102 a + 168$$

Factor

$$6\left(-a-7\right)\left(3a-4\right)$$

Solution Steps

Factor out 6.

$$6\left(-3a^{2}-17a+28\right)$$

Consider -3a^{2}-17a+28. Factor the expression by grouping. First, the expression needs to be rewritten as -3a^{2}+pa+qa+28. To find p and q, set up a system to be solved.

$$p+q=-17 pq=-3\times 28=-84$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -84.

$$1,-84 2,-42 3,-28 4,-21 6,-14 7,-12$$

Calculate the sum for each pair.

$$1-84=-83 2-42=-40 3-28=-25 4-21=-17 6-14=-8 7-12=-5$$

The solution is the pair that gives sum -17.

$$p=4 q=-21$$

Rewrite -3a^{2}-17a+28 as \left(-3a^{2}+4a\right)+\left(-21a+28\right).

$$\left(-3a^{2}+4a\right)+\left(-21a+28\right)$$

Factor out -a in the first and -7 in the second group.

$$-a\left(3a-4\right)-7\left(3a-4\right)$$

Factor out common term 3a-4 by using distributive property.

$$\left(3a-4\right)\left(-a-7\right)$$

Rewrite the complete factored expression.

$$6\left(3a-4\right)\left(-a-7\right)$$

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.

$$-18a^{2}-102a+168=0$$

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.

$$a=\frac{-\left(-102\right)±\sqrt{\left(-102\right)^{2}-4\left(-18\right)\times 168}}{2\left(-18\right)}$$

Square -102.

$$a=\frac{-\left(-102\right)±\sqrt{10404-4\left(-18\right)\times 168}}{2\left(-18\right)}$$

Multiply -4 times -18.

$$a=\frac{-\left(-102\right)±\sqrt{10404+72\times 168}}{2\left(-18\right)}$$

Multiply 72 times 168.

$$a=\frac{-\left(-102\right)±\sqrt{10404+12096}}{2\left(-18\right)}$$

Add 10404 to 12096.

$$a=\frac{-\left(-102\right)±\sqrt{22500}}{2\left(-18\right)}$$

Take the square root of 22500.

$$a=\frac{-\left(-102\right)±150}{2\left(-18\right)}$$

The opposite of -102 is 102.

$$a=\frac{102±150}{2\left(-18\right)}$$

Multiply 2 times -18.

$$a=\frac{102±150}{-36}$$

Now solve the equation a=\frac{102±150}{-36} when ± is plus. Add 102 to 150.

$$a=\frac{252}{-36}$$

Divide 252 by -36.

$$a=-7$$

Now solve the equation a=\frac{102±150}{-36} when ± is minus. Subtract 150 from 102.

$$a=-\frac{48}{-36}$$

Reduce the fraction \frac{-48}{-36} to lowest terms by extracting and canceling out 12.

$$a=\frac{4}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -7 for x_{1} and \frac{4}{3} for x_{2}.

$$-18a^{2}-102a+168=-18\left(a-\left(-7\right)\right)\left(a-\frac{4}{3}\right)$$

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

$$-18a^{2}-102a+168=-18\left(a+7\right)\left(a-\frac{4}{3}\right)$$

Subtract \frac{4}{3} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-18a^{2}-102a+168=-18\left(a+7\right)\times \frac{-3a+4}{-3}$$

Cancel out 3, the greatest common factor in -18 and 3.

$$-18a^{2}-102a+168=6\left(a+7\right)\left(-3a+4\right)$$

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.

$$x ^ 2 +\frac{17}{3}x -\frac{28}{3} = 0$$

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

$$r + s = -\frac{17}{3} rs = -\frac{28}{3}$$

Two numbers r and s sum up to -\frac{17}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{17}{3} = -\frac{17}{6}. 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>

$$r = -\frac{17}{6} - u s = -\frac{17}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{28}{3}

$$(-\frac{17}{6} - u) (-\frac{17}{6} + u) = -\frac{28}{3}$$

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

$$\frac{289}{36} - u^2 = -\frac{28}{3}$$

Simplify the expression by subtracting \frac{289}{36} on both sides

$$-u^2 = -\frac{28}{3}-\frac{289}{36} = -\frac{625}{36}$$

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

$$u^2 = \frac{625}{36} u = \pm\sqrt{\frac{625}{36}} = \pm \frac{25}{6}$$

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

$$r =-\frac{17}{6} - \frac{25}{6} = -7 s = -\frac{17}{6} + \frac{25}{6} = 1.333$$

Evaluate

$$168-102a-18a^{2}$$