Solve 21y^2-25y-4

$$21 y ^ { 2 } - 25 y - 4$$

Factor

$$\left(3y-4\right)\left(7y+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 21y^{2}+ay+by-4. To find a and b, set up a system to be solved.

$$a+b=-25 ab=21\left(-4\right)=-84$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -84.

$$1,-84 2,-42 3,-28 4,-21 6,-14 7,-12$$

Calculate the sum for each pair.

$$1-84=-83 2-42=-40 3-28=-25 4-21=-17 6-14=-8 7-12=-5$$

The solution is the pair that gives sum -25.

$$a=-28 b=3$$

Rewrite 21y^{2}-25y-4 as \left(21y^{2}-28y\right)+\left(3y-4\right).

$$\left(21y^{2}-28y\right)+\left(3y-4\right)$$

Factor out 7y in 21y^{2}-28y.

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

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

$$\left(3y-4\right)\left(7y+1\right)$$

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.

$$21y^{2}-25y-4=0$$

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.

$$y=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 21\left(-4\right)}}{2\times 21}$$

Square -25.

$$y=\frac{-\left(-25\right)±\sqrt{625-4\times 21\left(-4\right)}}{2\times 21}$$

Multiply -4 times 21.

$$y=\frac{-\left(-25\right)±\sqrt{625-84\left(-4\right)}}{2\times 21}$$

Multiply -84 times -4.

$$y=\frac{-\left(-25\right)±\sqrt{625+336}}{2\times 21}$$

Add 625 to 336.

$$y=\frac{-\left(-25\right)±\sqrt{961}}{2\times 21}$$

Take the square root of 961.

$$y=\frac{-\left(-25\right)±31}{2\times 21}$$

The opposite of -25 is 25.

$$y=\frac{25±31}{2\times 21}$$

Multiply 2 times 21.

$$y=\frac{25±31}{42}$$

Now solve the equation y=\frac{25±31}{42} when ± is plus. Add 25 to 31.

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

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

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

Now solve the equation y=\frac{25±31}{42} when ± is minus. Subtract 31 from 25.

$$y=-\frac{6}{42}$$

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

$$y=-\frac{1}{7}$$

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

$$21y^{2}-25y-4=21\left(y-\frac{4}{3}\right)\left(y-\left(-\frac{1}{7}\right)\right)$$

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

$$21y^{2}-25y-4=21\left(y-\frac{4}{3}\right)\left(y+\frac{1}{7}\right)$$

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

$$21y^{2}-25y-4=21\times \frac{3y-4}{3}\left(y+\frac{1}{7}\right)$$

Add \frac{1}{7} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$21y^{2}-25y-4=21\times \frac{3y-4}{3}\times \frac{7y+1}{7}$$

Multiply \frac{3y-4}{3} times \frac{7y+1}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$21y^{2}-25y-4=21\times \frac{\left(3y-4\right)\left(7y+1\right)}{3\times 7}$$

Multiply 3 times 7.

$$21y^{2}-25y-4=21\times \frac{\left(3y-4\right)\left(7y+1\right)}{21}$$

Cancel out 21, the greatest common factor in 21 and 21.

$$21y^{2}-25y-4=\left(3y-4\right)\left(7y+1\right)$$

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 21

$$x ^ 2 -\frac{25}{21}x -\frac{4}{21} = 0$$

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

$$r + s = \frac{25}{21} rs = -\frac{4}{21}$$

Two numbers r and s sum up to \frac{25}{21} exactly when the average of the two numbers is \frac{1}{2}*\frac{25}{21} = \frac{25}{42}. 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>

$$r = \frac{25}{42} - u s = \frac{25}{42} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{21}

$$(\frac{25}{42} - u) (\frac{25}{42} + u) = -\frac{4}{21}$$

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

$$\frac{625}{1764} - u^2 = -\frac{4}{21}$$

Simplify the expression by subtracting \frac{625}{1764} on both sides

$$-u^2 = -\frac{4}{21}-\frac{625}{1764} = -\frac{961}{1764}$$

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

$$u^2 = \frac{961}{1764} u = \pm\sqrt{\frac{961}{1764}} = \pm \frac{31}{42}$$

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

$$r =\frac{25}{42} - \frac{31}{42} = -0.143 s = \frac{25}{42} + \frac{31}{42} = 1.333$$

Evaluate

$$\left(3y-4\right)\left(7y+1\right)$$