Solve 25y^2-33y+8

$$25 y ^ { 2 } - 33 y + 8$$

Factor

$$\left(y-1\right)\left(25y-8\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 25y^{2}+ay+by+8. To find a and b, set up a system to be solved.

$$a+b=-33 ab=25\times 8=200$$

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 200.

$$-1,-200 -2,-100 -4,-50 -5,-40 -8,-25 -10,-20$$

Calculate the sum for each pair.

$$-1-200=-201 -2-100=-102 -4-50=-54 -5-40=-45 -8-25=-33 -10-20=-30$$

The solution is the pair that gives sum -33.

$$a=-25 b=-8$$

Rewrite 25y^{2}-33y+8 as \left(25y^{2}-25y\right)+\left(-8y+8\right).

$$\left(25y^{2}-25y\right)+\left(-8y+8\right)$$

Factor out 25y in the first and -8 in the second group.

$$25y\left(y-1\right)-8\left(y-1\right)$$

Factor out common term y-1 by using distributive property.

$$\left(y-1\right)\left(25y-8\right)$$

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.

$$25y^{2}-33y+8=0$$

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.

$$y=\frac{-\left(-33\right)±\sqrt{\left(-33\right)^{2}-4\times 25\times 8}}{2\times 25}$$

Square -33.

$$y=\frac{-\left(-33\right)±\sqrt{1089-4\times 25\times 8}}{2\times 25}$$

Multiply -4 times 25.

$$y=\frac{-\left(-33\right)±\sqrt{1089-100\times 8}}{2\times 25}$$

Multiply -100 times 8.

$$y=\frac{-\left(-33\right)±\sqrt{1089-800}}{2\times 25}$$

Add 1089 to -800.

$$y=\frac{-\left(-33\right)±\sqrt{289}}{2\times 25}$$

Take the square root of 289.

$$y=\frac{-\left(-33\right)±17}{2\times 25}$$

The opposite of -33 is 33.

$$y=\frac{33±17}{2\times 25}$$

Multiply 2 times 25.

$$y=\frac{33±17}{50}$$

Now solve the equation y=\frac{33±17}{50} when ± is plus. Add 33 to 17.

$$y=\frac{50}{50}$$

Divide 50 by 50.

$$y=1$$

Now solve the equation y=\frac{33±17}{50} when ± is minus. Subtract 17 from 33.

$$y=\frac{16}{50}$$

Reduce the fraction \frac{16}{50} to lowest terms by extracting and canceling out 2.

$$y=\frac{8}{25}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and \frac{8}{25} for x_{2}.

$$25y^{2}-33y+8=25\left(y-1\right)\left(y-\frac{8}{25}\right)$$

Subtract \frac{8}{25} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$25y^{2}-33y+8=25\left(y-1\right)\times \frac{25y-8}{25}$$

Cancel out 25, the greatest common factor in 25 and 25.

$$25y^{2}-33y+8=\left(y-1\right)\left(25y-8\right)$$

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 25

$$x ^ 2 -\frac{33}{25}x +\frac{8}{25} = 0$$

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

$$r + s = \frac{33}{25} rs = \frac{8}{25}$$

Two numbers r and s sum up to \frac{33}{25} exactly when the average of the two numbers is \frac{1}{2}*\frac{33}{25} = \frac{33}{50}. 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>

$$r = \frac{33}{50} - u s = \frac{33}{50} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{8}{25}

$$(\frac{33}{50} - u) (\frac{33}{50} + u) = \frac{8}{25}$$

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

$$\frac{1089}{2500} - u^2 = \frac{8}{25}$$

Simplify the expression by subtracting \frac{1089}{2500} on both sides

$$-u^2 = \frac{8}{25}-\frac{1089}{2500} = -\frac{289}{2500}$$

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

$$u^2 = \frac{289}{2500} u = \pm\sqrt{\frac{289}{2500}} = \pm \frac{17}{50}$$

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

$$r =\frac{33}{50} - \frac{17}{50} = 0.320 s = \frac{33}{50} + \frac{17}{50} = 1$$

Evaluate

$$\left(y-1\right)\left(25y-8\right)$$