Solve 11a^2-70a+24

$$11 a ^ { 2 } - 70 a + 24$$

Factor

$$\left(a-6\right)\left(11a-4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 11a^{2}+pa+qa+24. To find p and q, set up a system to be solved.

$$p+q=-70 pq=11\times 24=264$$

Since pq is positive, p and q have the same sign. Since p+q is negative, p and q are both negative. List all such integer pairs that give product 264.

$$-1,-264 -2,-132 -3,-88 -4,-66 -6,-44 -8,-33 -11,-24 -12,-22$$

Calculate the sum for each pair.

$$-1-264=-265 -2-132=-134 -3-88=-91 -4-66=-70 -6-44=-50 -8-33=-41 -11-24=-35 -12-22=-34$$

The solution is the pair that gives sum -70.

$$p=-66 q=-4$$

Rewrite 11a^{2}-70a+24 as \left(11a^{2}-66a\right)+\left(-4a+24\right).

$$\left(11a^{2}-66a\right)+\left(-4a+24\right)$$

Factor out 11a in the first and -4 in the second group.

$$11a\left(a-6\right)-4\left(a-6\right)$$

Factor out common term a-6 by using distributive property.

$$\left(a-6\right)\left(11a-4\right)$$

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.

$$11a^{2}-70a+24=0$$

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.

$$a=\frac{-\left(-70\right)±\sqrt{\left(-70\right)^{2}-4\times 11\times 24}}{2\times 11}$$

Square -70.

$$a=\frac{-\left(-70\right)±\sqrt{4900-4\times 11\times 24}}{2\times 11}$$

Multiply -4 times 11.

$$a=\frac{-\left(-70\right)±\sqrt{4900-44\times 24}}{2\times 11}$$

Multiply -44 times 24.

$$a=\frac{-\left(-70\right)±\sqrt{4900-1056}}{2\times 11}$$

Add 4900 to -1056.

$$a=\frac{-\left(-70\right)±\sqrt{3844}}{2\times 11}$$

Take the square root of 3844.

$$a=\frac{-\left(-70\right)±62}{2\times 11}$$

The opposite of -70 is 70.

$$a=\frac{70±62}{2\times 11}$$

Multiply 2 times 11.

$$a=\frac{70±62}{22}$$

Now solve the equation a=\frac{70±62}{22} when ± is plus. Add 70 to 62.

$$a=\frac{132}{22}$$

Divide 132 by 22.

$$a=6$$

Now solve the equation a=\frac{70±62}{22} when ± is minus. Subtract 62 from 70.

$$a=\frac{8}{22}$$

Reduce the fraction \frac{8}{22} to lowest terms by extracting and canceling out 2.

$$a=\frac{4}{11}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 6 for x_{1} and \frac{4}{11} for x_{2}.

$$11a^{2}-70a+24=11\left(a-6\right)\left(a-\frac{4}{11}\right)$$

Subtract \frac{4}{11} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$11a^{2}-70a+24=11\left(a-6\right)\times \frac{11a-4}{11}$$

Cancel out 11, the greatest common factor in 11 and 11.

$$11a^{2}-70a+24=\left(a-6\right)\left(11a-4\right)$$

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.This is achieved by dividing both sides of the equation by 11

$$x ^ 2 -\frac{70}{11}x +\frac{24}{11} = 0$$

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

$$r + s = \frac{70}{11} rs = \frac{24}{11}$$

Two numbers r and s sum up to \frac{70}{11} exactly when the average of the two numbers is \frac{1}{2}*\frac{70}{11} = \frac{35}{11}. 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>

$$r = \frac{35}{11} - u s = \frac{35}{11} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{24}{11}

$$(\frac{35}{11} - u) (\frac{35}{11} + u) = \frac{24}{11}$$

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

$$\frac{1225}{121} - u^2 = \frac{24}{11}$$

Simplify the expression by subtracting \frac{1225}{121} on both sides

$$-u^2 = \frac{24}{11}-\frac{1225}{121} = -\frac{961}{121}$$

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

$$u^2 = \frac{961}{121} u = \pm\sqrt{\frac{961}{121}} = \pm \frac{31}{11}$$

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

$$r =\frac{35}{11} - \frac{31}{11} = 0.364 s = \frac{35}{11} + \frac{31}{11} = 6$$

Evaluate

$$\left(a-6\right)\left(11a-4\right)$$