Solve 28y^2-29y+6

$$28 y ^ { 2 } - 29 y + 6$$

Factor

$$\left(4y-3\right)\left(7y-2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 28y^{2}+ay+by+6. To find a and b, set up a system to be solved.

$$a+b=-29 ab=28\times 6=168$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 168.

$$-1,-168 -2,-84 -3,-56 -4,-42 -6,-28 -7,-24 -8,-21 -12,-14$$

Calculate the sum for each pair.

$$-1-168=-169 -2-84=-86 -3-56=-59 -4-42=-46 -6-28=-34 -7-24=-31 -8-21=-29 -12-14=-26$$

The solution is the pair that gives sum -29.

$$a=-21 b=-8$$

Rewrite 28y^{2}-29y+6 as \left(28y^{2}-21y\right)+\left(-8y+6\right).

$$\left(28y^{2}-21y\right)+\left(-8y+6\right)$$

Factor out 7y in the first and -2 in the second group.

$$7y\left(4y-3\right)-2\left(4y-3\right)$$

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

$$\left(4y-3\right)\left(7y-2\right)$$

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.

$$28y^{2}-29y+6=0$$

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.

$$y=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 28\times 6}}{2\times 28}$$

Square -29.

$$y=\frac{-\left(-29\right)±\sqrt{841-4\times 28\times 6}}{2\times 28}$$

Multiply -4 times 28.

$$y=\frac{-\left(-29\right)±\sqrt{841-112\times 6}}{2\times 28}$$

Multiply -112 times 6.

$$y=\frac{-\left(-29\right)±\sqrt{841-672}}{2\times 28}$$

Add 841 to -672.

$$y=\frac{-\left(-29\right)±\sqrt{169}}{2\times 28}$$

Take the square root of 169.

$$y=\frac{-\left(-29\right)±13}{2\times 28}$$

The opposite of -29 is 29.

$$y=\frac{29±13}{2\times 28}$$

Multiply 2 times 28.

$$y=\frac{29±13}{56}$$

Now solve the equation y=\frac{29±13}{56} when ± is plus. Add 29 to 13.

$$y=\frac{42}{56}$$

Reduce the fraction \frac{42}{56} to lowest terms by extracting and canceling out 14.

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

Now solve the equation y=\frac{29±13}{56} when ± is minus. Subtract 13 from 29.

$$y=\frac{16}{56}$$

Reduce the fraction \frac{16}{56} to lowest terms by extracting and canceling out 8.

$$y=\frac{2}{7}$$

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

$$28y^{2}-29y+6=28\left(y-\frac{3}{4}\right)\left(y-\frac{2}{7}\right)$$

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

$$28y^{2}-29y+6=28\times \frac{4y-3}{4}\left(y-\frac{2}{7}\right)$$

Subtract \frac{2}{7} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$28y^{2}-29y+6=28\times \frac{4y-3}{4}\times \frac{7y-2}{7}$$

Multiply \frac{4y-3}{4} times \frac{7y-2}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$28y^{2}-29y+6=28\times \frac{\left(4y-3\right)\left(7y-2\right)}{4\times 7}$$

Multiply 4 times 7.

$$28y^{2}-29y+6=28\times \frac{\left(4y-3\right)\left(7y-2\right)}{28}$$

Cancel out 28, the greatest common factor in 28 and 28.

$$28y^{2}-29y+6=\left(4y-3\right)\left(7y-2\right)$$

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 28

$$x ^ 2 -\frac{29}{28}x +\frac{3}{14} = 0$$

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

$$r + s = \frac{29}{28} rs = \frac{3}{14}$$

Two numbers r and s sum up to \frac{29}{28} exactly when the average of the two numbers is \frac{1}{2}*\frac{29}{28} = \frac{29}{56}. 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>

$$r = \frac{29}{56} - u s = \frac{29}{56} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{14}

$$(\frac{29}{56} - u) (\frac{29}{56} + u) = \frac{3}{14}$$

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

$$\frac{841}{3136} - u^2 = \frac{3}{14}$$

Simplify the expression by subtracting \frac{841}{3136} on both sides

$$-u^2 = \frac{3}{14}-\frac{841}{3136} = -\frac{169}{3136}$$

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

$$u^2 = \frac{169}{3136} u = \pm\sqrt{\frac{169}{3136}} = \pm \frac{13}{56}$$

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

$$r =\frac{29}{56} - \frac{13}{56} = 0.286 s = \frac{29}{56} + \frac{13}{56} = 0.750$$

Evaluate

$$\left(4y-3\right)\left(7y-2\right)$$