Solve 35a^2-275a-360

$$35 a ^ { 2 } - 275 a - 360$$

Factor

$$5\left(a-9\right)\left(7a+8\right)$$

Solution Steps

Factor out 5.

$$5\left(7a^{2}-55a-72\right)$$

Consider 7a^{2}-55a-72. Factor the expression by grouping. First, the expression needs to be rewritten as 7a^{2}+pa+qa-72. To find p and q, set up a system to be solved.

$$p+q=-55 pq=7\left(-72\right)=-504$$

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 -504.

$$1,-504 2,-252 3,-168 4,-126 6,-84 7,-72 8,-63 9,-56 12,-42 14,-36 18,-28 21,-24$$

Calculate the sum for each pair.

$$1-504=-503 2-252=-250 3-168=-165 4-126=-122 6-84=-78 7-72=-65 8-63=-55 9-56=-47 12-42=-30 14-36=-22 18-28=-10 21-24=-3$$

The solution is the pair that gives sum -55.

$$p=-63 q=8$$

Rewrite 7a^{2}-55a-72 as \left(7a^{2}-63a\right)+\left(8a-72\right).

$$\left(7a^{2}-63a\right)+\left(8a-72\right)$$

Factor out 7a in the first and 8 in the second group.

$$7a\left(a-9\right)+8\left(a-9\right)$$

Factor out common term a-9 by using distributive property.

$$\left(a-9\right)\left(7a+8\right)$$

Rewrite the complete factored expression.

$$5\left(a-9\right)\left(7a+8\right)$$

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.

$$35a^{2}-275a-360=0$$

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.

$$a=\frac{-\left(-275\right)±\sqrt{\left(-275\right)^{2}-4\times 35\left(-360\right)}}{2\times 35}$$

Square -275.

$$a=\frac{-\left(-275\right)±\sqrt{75625-4\times 35\left(-360\right)}}{2\times 35}$$

Multiply -4 times 35.

$$a=\frac{-\left(-275\right)±\sqrt{75625-140\left(-360\right)}}{2\times 35}$$

Multiply -140 times -360.

$$a=\frac{-\left(-275\right)±\sqrt{75625+50400}}{2\times 35}$$

Add 75625 to 50400.

$$a=\frac{-\left(-275\right)±\sqrt{126025}}{2\times 35}$$

Take the square root of 126025.

$$a=\frac{-\left(-275\right)±355}{2\times 35}$$

The opposite of -275 is 275.

$$a=\frac{275±355}{2\times 35}$$

Multiply 2 times 35.

$$a=\frac{275±355}{70}$$

Now solve the equation a=\frac{275±355}{70} when ± is plus. Add 275 to 355.

$$a=\frac{630}{70}$$

Divide 630 by 70.

$$a=9$$

Now solve the equation a=\frac{275±355}{70} when ± is minus. Subtract 355 from 275.

$$a=-\frac{80}{70}$$

Reduce the fraction \frac{-80}{70} to lowest terms by extracting and canceling out 10.

$$a=-\frac{8}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 9 for x_{1} and -\frac{8}{7} for x_{2}.

$$35a^{2}-275a-360=35\left(a-9\right)\left(a-\left(-\frac{8}{7}\right)\right)$$

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

$$35a^{2}-275a-360=35\left(a-9\right)\left(a+\frac{8}{7}\right)$$

Add \frac{8}{7} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$35a^{2}-275a-360=35\left(a-9\right)\times \frac{7a+8}{7}$$

Cancel out 7, the greatest common factor in 35 and 7.

$$35a^{2}-275a-360=5\left(a-9\right)\left(7a+8\right)$$

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 35

$$x ^ 2 -\frac{55}{7}x -\frac{72}{7} = 0$$

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

$$r + s = \frac{55}{7} rs = -\frac{72}{7}$$

Two numbers r and s sum up to \frac{55}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{55}{7} = \frac{55}{14}. 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>

$$r = \frac{55}{14} - u s = \frac{55}{14} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{72}{7}

$$(\frac{55}{14} - u) (\frac{55}{14} + u) = -\frac{72}{7}$$

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

$$\frac{3025}{196} - u^2 = -\frac{72}{7}$$

Simplify the expression by subtracting \frac{3025}{196} on both sides

$$-u^2 = -\frac{72}{7}-\frac{3025}{196} = -\frac{5041}{196}$$

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

$$u^2 = \frac{5041}{196} u = \pm\sqrt{\frac{5041}{196}} = \pm \frac{71}{14}$$

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

$$r =\frac{55}{14} - \frac{71}{14} = -1.143 s = \frac{55}{14} + \frac{71}{14} = 9$$

Evaluate

$$5\left(a-9\right)\left(7a+8\right)$$