Solve 10x^2+7x-3

$$10 x ^ { 2 } + 7 x - 3$$

Factor

$$\left(10x-3\right)\left(x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=7 ab=10\left(-3\right)=-30$$

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 -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 7.

$$a=-3 b=10$$

Rewrite 10x^{2}+7x-3 as \left(10x^{2}-3x\right)+\left(10x-3\right).

$$\left(10x^{2}-3x\right)+\left(10x-3\right)$$

Factor out x in 10x^{2}-3x.

$$x\left(10x-3\right)+10x-3$$

Factor out common term 10x-3 by using distributive property.

$$\left(10x-3\right)\left(x+1\right)$$

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.

$$10x^{2}+7x-3=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\times 10\left(-3\right)}}{2\times 10}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 10\left(-3\right)}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-7±\sqrt{49-40\left(-3\right)}}{2\times 10}$$

Multiply -40 times -3.

$$x=\frac{-7±\sqrt{49+120}}{2\times 10}$$

Add 49 to 120.

$$x=\frac{-7±\sqrt{169}}{2\times 10}$$

Take the square root of 169.

$$x=\frac{-7±13}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-7±13}{20}$$

Now solve the equation x=\frac{-7±13}{20} when ± is plus. Add -7 to 13.

$$x=\frac{6}{20}$$

Reduce the fraction \frac{6}{20} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{10}$$

Now solve the equation x=\frac{-7±13}{20} when ± is minus. Subtract 13 from -7.

$$x=-\frac{20}{20}$$

Divide -20 by 20.

$$x=-1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{10} for x_{1} and -1 for x_{2}.

$$10x^{2}+7x-3=10\left(x-\frac{3}{10}\right)\left(x-\left(-1\right)\right)$$

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

$$10x^{2}+7x-3=10\left(x-\frac{3}{10}\right)\left(x+1\right)$$

Subtract \frac{3}{10} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10x^{2}+7x-3=10\times \frac{10x-3}{10}\left(x+1\right)$$

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

$$10x^{2}+7x-3=\left(10x-3\right)\left(x+1\right)$$

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 10

$$x ^ 2 +\frac{7}{10}x -\frac{3}{10} = 0$$

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

$$r + s = -\frac{7}{10} rs = -\frac{3}{10}$$

Two numbers r and s sum up to -\frac{7}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{10} = -\frac{7}{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>

$$r = -\frac{7}{20} - u s = -\frac{7}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{10}

$$(-\frac{7}{20} - u) (-\frac{7}{20} + u) = -\frac{3}{10}$$

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

$$\frac{49}{400} - u^2 = -\frac{3}{10}$$

Simplify the expression by subtracting \frac{49}{400} on both sides

$$-u^2 = -\frac{3}{10}-\frac{49}{400} = -\frac{169}{400}$$

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

$$u^2 = \frac{169}{400} u = \pm\sqrt{\frac{169}{400}} = \pm \frac{13}{20}$$

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

$$r =-\frac{7}{20} - \frac{13}{20} = -1 s = -\frac{7}{20} + \frac{13}{20} = 0.300$$

Evaluate

$$\left(10x-3\right)\left(x+1\right)$$