Solve 12b^2+13b-35

$$12 b ^ { 2 } + 13 b - 35$$

Factor

$$\left(4b-5\right)\left(3b+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12b^{2}+pb+qb-35. To find p and q, set up a system to be solved.

$$p+q=13 pq=12\left(-35\right)=-420$$

Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -420.

$$-1,420 -2,210 -3,140 -4,105 -5,84 -6,70 -7,60 -10,42 -12,35 -14,30 -15,28 -20,21$$

Calculate the sum for each pair.

$$-1+420=419 -2+210=208 -3+140=137 -4+105=101 -5+84=79 -6+70=64 -7+60=53 -10+42=32 -12+35=23 -14+30=16 -15+28=13 -20+21=1$$

The solution is the pair that gives sum 13.

$$p=-15 q=28$$

Rewrite 12b^{2}+13b-35 as \left(12b^{2}-15b\right)+\left(28b-35\right).

$$\left(12b^{2}-15b\right)+\left(28b-35\right)$$

Factor out 3b in the first and 7 in the second group.

$$3b\left(4b-5\right)+7\left(4b-5\right)$$

Factor out common term 4b-5 by using distributive property.

$$\left(4b-5\right)\left(3b+7\right)$$

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.

$$12b^{2}+13b-35=0$$

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.

$$b=\frac{-13±\sqrt{13^{2}-4\times 12\left(-35\right)}}{2\times 12}$$

Square 13.

$$b=\frac{-13±\sqrt{169-4\times 12\left(-35\right)}}{2\times 12}$$

Multiply -4 times 12.

$$b=\frac{-13±\sqrt{169-48\left(-35\right)}}{2\times 12}$$

Multiply -48 times -35.

$$b=\frac{-13±\sqrt{169+1680}}{2\times 12}$$

Add 169 to 1680.

$$b=\frac{-13±\sqrt{1849}}{2\times 12}$$

Take the square root of 1849.

$$b=\frac{-13±43}{2\times 12}$$

Multiply 2 times 12.

$$b=\frac{-13±43}{24}$$

Now solve the equation b=\frac{-13±43}{24} when ± is plus. Add -13 to 43.

$$b=\frac{30}{24}$$

Reduce the fraction \frac{30}{24} to lowest terms by extracting and canceling out 6.

$$b=\frac{5}{4}$$

Now solve the equation b=\frac{-13±43}{24} when ± is minus. Subtract 43 from -13.

$$b=-\frac{56}{24}$$

Reduce the fraction \frac{-56}{24} to lowest terms by extracting and canceling out 8.

$$b=-\frac{7}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{4} for x_{1} and -\frac{7}{3} for x_{2}.

$$12b^{2}+13b-35=12\left(b-\frac{5}{4}\right)\left(b-\left(-\frac{7}{3}\right)\right)$$

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

$$12b^{2}+13b-35=12\left(b-\frac{5}{4}\right)\left(b+\frac{7}{3}\right)$$

Subtract \frac{5}{4} from b by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12b^{2}+13b-35=12\times \frac{4b-5}{4}\left(b+\frac{7}{3}\right)$$

Add \frac{7}{3} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12b^{2}+13b-35=12\times \frac{4b-5}{4}\times \frac{3b+7}{3}$$

Multiply \frac{4b-5}{4} times \frac{3b+7}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12b^{2}+13b-35=12\times \frac{\left(4b-5\right)\left(3b+7\right)}{4\times 3}$$

Multiply 4 times 3.

$$12b^{2}+13b-35=12\times \frac{\left(4b-5\right)\left(3b+7\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12b^{2}+13b-35=\left(4b-5\right)\left(3b+7\right)$$

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 12

$$x ^ 2 +\frac{13}{12}x -\frac{35}{12} = 0$$

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

$$r + s = -\frac{13}{12} rs = -\frac{35}{12}$$

Two numbers r and s sum up to -\frac{13}{12} exactly when the average of the two numbers is \frac{1}{2}*-\frac{13}{12} = -\frac{13}{24}. 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>

$$r = -\frac{13}{24} - u s = -\frac{13}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{35}{12}

$$(-\frac{13}{24} - u) (-\frac{13}{24} + u) = -\frac{35}{12}$$

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

$$\frac{169}{576} - u^2 = -\frac{35}{12}$$

Simplify the expression by subtracting \frac{169}{576} on both sides

$$-u^2 = -\frac{35}{12}-\frac{169}{576} = -\frac{1849}{576}$$

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

$$u^2 = \frac{1849}{576} u = \pm\sqrt{\frac{1849}{576}} = \pm \frac{43}{24}$$

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

$$r =-\frac{13}{24} - \frac{43}{24} = -2.333 s = -\frac{13}{24} + \frac{43}{24} = 1.250$$

Evaluate

$$\left(4b-5\right)\left(3b+7\right)$$