Solve 20x^2+51x+28

$$20 x ^ { 2 } + 51 x + 28$$

Factor

$$\left(4x+7\right)\left(5x+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20x^{2}+ax+bx+28. To find a and b, set up a system to be solved.

$$a+b=51 ab=20\times 28=560$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 560.

$$1,560 2,280 4,140 5,112 7,80 8,70 10,56 14,40 16,35 20,28$$

Calculate the sum for each pair.

$$1+560=561 2+280=282 4+140=144 5+112=117 7+80=87 8+70=78 10+56=66 14+40=54 16+35=51 20+28=48$$

The solution is the pair that gives sum 51.

$$a=16 b=35$$

Rewrite 20x^{2}+51x+28 as \left(20x^{2}+16x\right)+\left(35x+28\right).

$$\left(20x^{2}+16x\right)+\left(35x+28\right)$$

Factor out 4x in the first and 7 in the second group.

$$4x\left(5x+4\right)+7\left(5x+4\right)$$

Factor out common term 5x+4 by using distributive property.

$$\left(5x+4\right)\left(4x+7\right)$$

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.

$$20x^{2}+51x+28=0$$

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.

$$x=\frac{-51±\sqrt{51^{2}-4\times 20\times 28}}{2\times 20}$$

Square 51.

$$x=\frac{-51±\sqrt{2601-4\times 20\times 28}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-51±\sqrt{2601-80\times 28}}{2\times 20}$$

Multiply -80 times 28.

$$x=\frac{-51±\sqrt{2601-2240}}{2\times 20}$$

Add 2601 to -2240.

$$x=\frac{-51±\sqrt{361}}{2\times 20}$$

Take the square root of 361.

$$x=\frac{-51±19}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-51±19}{40}$$

Now solve the equation x=\frac{-51±19}{40} when ± is plus. Add -51 to 19.

$$x=-\frac{32}{40}$$

Reduce the fraction \frac{-32}{40} to lowest terms by extracting and canceling out 8.

$$x=-\frac{4}{5}$$

Now solve the equation x=\frac{-51±19}{40} when ± is minus. Subtract 19 from -51.

$$x=-\frac{70}{40}$$

Reduce the fraction \frac{-70}{40} to lowest terms by extracting and canceling out 10.

$$x=-\frac{7}{4}$$

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

$$20x^{2}+51x+28=20\left(x-\left(-\frac{4}{5}\right)\right)\left(x-\left(-\frac{7}{4}\right)\right)$$

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

$$20x^{2}+51x+28=20\left(x+\frac{4}{5}\right)\left(x+\frac{7}{4}\right)$$

Add \frac{4}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+51x+28=20\times \frac{5x+4}{5}\left(x+\frac{7}{4}\right)$$

Add \frac{7}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20x^{2}+51x+28=20\times \frac{5x+4}{5}\times \frac{4x+7}{4}$$

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

$$20x^{2}+51x+28=20\times \frac{\left(5x+4\right)\left(4x+7\right)}{5\times 4}$$

Multiply 5 times 4.

$$20x^{2}+51x+28=20\times \frac{\left(5x+4\right)\left(4x+7\right)}{20}$$

Cancel out 20, the greatest common factor in 20 and 20.

$$20x^{2}+51x+28=\left(5x+4\right)\left(4x+7\right)$$

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 20

$$x ^ 2 +\frac{51}{20}x +\frac{7}{5} = 0$$

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

$$r + s = -\frac{51}{20} rs = \frac{7}{5}$$

Two numbers r and s sum up to -\frac{51}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{51}{20} = -\frac{51}{40}. 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>

$$r = -\frac{51}{40} - u s = -\frac{51}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{7}{5}

$$(-\frac{51}{40} - u) (-\frac{51}{40} + u) = \frac{7}{5}$$

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

$$\frac{2601}{1600} - u^2 = \frac{7}{5}$$

Simplify the expression by subtracting \frac{2601}{1600} on both sides

$$-u^2 = \frac{7}{5}-\frac{2601}{1600} = -\frac{361}{1600}$$

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

$$u^2 = \frac{361}{1600} u = \pm\sqrt{\frac{361}{1600}} = \pm \frac{19}{40}$$

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

$$r =-\frac{51}{40} - \frac{19}{40} = -1.750 s = -\frac{51}{40} + \frac{19}{40} = -0.800$$

Evaluate

$$\left(4x+7\right)\left(5x+4\right)$$