Solve 10y^2+29y-21

$$10 y ^ { 2 } + 29 y - 21$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10y^{2}+ay+by-21. To find a and b, set up a system to be solved.

$$a+b=29 ab=10\left(-21\right)=-210$$

Since ab is negative, a and b have the opposite signs. Since a+b 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 29.

$$a=-6 b=35$$

Rewrite 10y^{2}+29y-21 as \left(10y^{2}-6y\right)+\left(35y-21\right).

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

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

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

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

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

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.

$$10y^{2}+29y-21=0$$

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.

$$y=\frac{-29±\sqrt{29^{2}-4\times 10\left(-21\right)}}{2\times 10}$$

Square 29.

$$y=\frac{-29±\sqrt{841-4\times 10\left(-21\right)}}{2\times 10}$$

Multiply -4 times 10.

$$y=\frac{-29±\sqrt{841-40\left(-21\right)}}{2\times 10}$$

Multiply -40 times -21.

$$y=\frac{-29±\sqrt{841+840}}{2\times 10}$$

Add 841 to 840.

$$y=\frac{-29±\sqrt{1681}}{2\times 10}$$

Take the square root of 1681.

$$y=\frac{-29±41}{2\times 10}$$

Multiply 2 times 10.

$$y=\frac{-29±41}{20}$$

Now solve the equation y=\frac{-29±41}{20} when ± is plus. Add -29 to 41.

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

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

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

Now solve the equation y=\frac{-29±41}{20} when ± is minus. Subtract 41 from -29.

$$y=-\frac{70}{20}$$

Reduce the fraction \frac{-70}{20} to lowest terms by extracting and canceling out 10.

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

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

$$10y^{2}+29y-21=10\left(y-\frac{3}{5}\right)\left(y-\left(-\frac{7}{2}\right)\right)$$

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

$$10y^{2}+29y-21=10\left(y-\frac{3}{5}\right)\left(y+\frac{7}{2}\right)$$

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

$$10y^{2}+29y-21=10\times \frac{5y-3}{5}\left(y+\frac{7}{2}\right)$$

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

$$10y^{2}+29y-21=10\times \frac{5y-3}{5}\times \frac{2y+7}{2}$$

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

$$10y^{2}+29y-21=10\times \frac{\left(5y-3\right)\left(2y+7\right)}{5\times 2}$$

Multiply 5 times 2.

$$10y^{2}+29y-21=10\times \frac{\left(5y-3\right)\left(2y+7\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10y^{2}+29y-21=\left(5y-3\right)\left(2y+7\right)$$

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 10

$$x ^ 2 +\frac{29}{10}x -\frac{21}{10} = 0$$

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

$$r + s = -\frac{29}{10} rs = -\frac{21}{10}$$

Two numbers r and s sum up to -\frac{29}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{29}{10} = -\frac{29}{20}. 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>

$$r = -\frac{29}{20} - u s = -\frac{29}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{21}{10}

$$(-\frac{29}{20} - u) (-\frac{29}{20} + u) = -\frac{21}{10}$$

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

$$\frac{841}{400} - u^2 = -\frac{21}{10}$$

Simplify the expression by subtracting \frac{841}{400} on both sides

$$-u^2 = -\frac{21}{10}-\frac{841}{400} = -\frac{1681}{400}$$

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

$$u^2 = \frac{1681}{400} u = \pm\sqrt{\frac{1681}{400}} = \pm \frac{41}{20}$$

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

$$r =-\frac{29}{20} - \frac{41}{20} = -3.500 s = -\frac{29}{20} + \frac{41}{20} = 0.600$$

Evaluate

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