Solve 6b^2+11b-35

$$6 b ^ { 2 } + 11 b - 35$$

Factor

$$\left(3b-5\right)\left(2b+7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 6b^{2}+pb+qb-35. To find p and q, set up a system to be solved.

$$p+q=11 pq=6\left(-35\right)=-210$$

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 -210.

$$-1,210 -2,105 -3,70 -5,42 -6,35 -7,30 -10,21 -14,15$$

Calculate the sum for each pair.

$$-1+210=209 -2+105=103 -3+70=67 -5+42=37 -6+35=29 -7+30=23 -10+21=11 -14+15=1$$

The solution is the pair that gives sum 11.

$$p=-10 q=21$$

Rewrite 6b^{2}+11b-35 as \left(6b^{2}-10b\right)+\left(21b-35\right).

$$\left(6b^{2}-10b\right)+\left(21b-35\right)$$

Factor out 2b in the first and 7 in the second group.

$$2b\left(3b-5\right)+7\left(3b-5\right)$$

Factor out common term 3b-5 by using distributive property.

$$\left(3b-5\right)\left(2b+7\right)$$

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.

$$6b^{2}+11b-35=0$$

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.

$$b=\frac{-11±\sqrt{11^{2}-4\times 6\left(-35\right)}}{2\times 6}$$

Square 11.

$$b=\frac{-11±\sqrt{121-4\times 6\left(-35\right)}}{2\times 6}$$

Multiply -4 times 6.

$$b=\frac{-11±\sqrt{121-24\left(-35\right)}}{2\times 6}$$

Multiply -24 times -35.

$$b=\frac{-11±\sqrt{121+840}}{2\times 6}$$

Add 121 to 840.

$$b=\frac{-11±\sqrt{961}}{2\times 6}$$

Take the square root of 961.

$$b=\frac{-11±31}{2\times 6}$$

Multiply 2 times 6.

$$b=\frac{-11±31}{12}$$

Now solve the equation b=\frac{-11±31}{12} when ± is plus. Add -11 to 31.

$$b=\frac{20}{12}$$

Reduce the fraction \frac{20}{12} to lowest terms by extracting and canceling out 4.

$$b=\frac{5}{3}$$

Now solve the equation b=\frac{-11±31}{12} when ± is minus. Subtract 31 from -11.

$$b=-\frac{42}{12}$$

Reduce the fraction \frac{-42}{12} to lowest terms by extracting and canceling out 6.

$$b=-\frac{7}{2}$$

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

$$6b^{2}+11b-35=6\left(b-\frac{5}{3}\right)\left(b-\left(-\frac{7}{2}\right)\right)$$

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

$$6b^{2}+11b-35=6\left(b-\frac{5}{3}\right)\left(b+\frac{7}{2}\right)$$

Subtract \frac{5}{3} from b by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$6b^{2}+11b-35=6\times \frac{3b-5}{3}\left(b+\frac{7}{2}\right)$$

Add \frac{7}{2} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$6b^{2}+11b-35=6\times \frac{3b-5}{3}\times \frac{2b+7}{2}$$

Multiply \frac{3b-5}{3} times \frac{2b+7}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$6b^{2}+11b-35=6\times \frac{\left(3b-5\right)\left(2b+7\right)}{3\times 2}$$

Multiply 3 times 2.

$$6b^{2}+11b-35=6\times \frac{\left(3b-5\right)\left(2b+7\right)}{6}$$

Cancel out 6, the greatest common factor in 6 and 6.

$$6b^{2}+11b-35=\left(3b-5\right)\left(2b+7\right)$$

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 6

$$x ^ 2 +\frac{11}{6}x -\frac{35}{6} = 0$$

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

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

Two numbers r and s sum up to -\frac{11}{6} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{6} = -\frac{11}{12}. 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>

$$r = -\frac{11}{12} - u s = -\frac{11}{12} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{35}{6}

$$(-\frac{11}{12} - u) (-\frac{11}{12} + u) = -\frac{35}{6}$$

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

$$\frac{121}{144} - u^2 = -\frac{35}{6}$$

Simplify the expression by subtracting \frac{121}{144} on both sides

$$-u^2 = -\frac{35}{6}-\frac{121}{144} = -\frac{961}{144}$$

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

$$u^2 = \frac{961}{144} u = \pm\sqrt{\frac{961}{144}} = \pm \frac{31}{12}$$

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

$$r =-\frac{11}{12} - \frac{31}{12} = -3.500 s = -\frac{11}{12} + \frac{31}{12} = 1.667$$

Evaluate

$$\left(3b-5\right)\left(2b+7\right)$$