Solve -3x^2-2x+5)-(-2x^2+7x+2)

$$- 3 x ^ { 2 } - 2 x + 5 ) - ( - 2 x ^ { 2 } + 7 x + 2 )$$

Factor

$$\left(1-x\right)\left(3x+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3\times 5=-15$$

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 -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=3 b=-5$$

Rewrite -3x^{2}-2x+5 as \left(-3x^{2}+3x\right)+\left(-5x+5\right).

$$\left(-3x^{2}+3x\right)+\left(-5x+5\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(-x+1\right)+5\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(3x+5\right)$$

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.

$$-3x^{2}-2x+5=0$$

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.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-3\right)\times 5}}{2\left(-3\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-3\right)\times 5}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+12\times 5}}{2\left(-3\right)}$$

Multiply 12 times 5.

$$x=\frac{-\left(-2\right)±\sqrt{4+60}}{2\left(-3\right)}$$

Add 4 to 60.

$$x=\frac{-\left(-2\right)±\sqrt{64}}{2\left(-3\right)}$$

Take the square root of 64.

$$x=\frac{-\left(-2\right)±8}{2\left(-3\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±8}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{2±8}{-6}$$

Now solve the equation x=\frac{2±8}{-6} when ± is plus. Add 2 to 8.

$$x=\frac{10}{-6}$$

Reduce the fraction \frac{10}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{3}$$

Now solve the equation x=\frac{2±8}{-6} when ± is minus. Subtract 8 from 2.

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

Divide -6 by -6.

$$x=1$$

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

$$-3x^{2}-2x+5=-3\left(x-\left(-\frac{5}{3}\right)\right)\left(x-1\right)$$

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

$$-3x^{2}-2x+5=-3\left(x+\frac{5}{3}\right)\left(x-1\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-3x^{2}-2x+5=-3\times \frac{-3x-5}{-3}\left(x-1\right)$$

Cancel out 3, the greatest common factor in -3 and 3.

$$-3x^{2}-2x+5=\left(-3x-5\right)\left(x-1\right)$$

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.

$$x ^ 2 +\frac{2}{3}x -\frac{5}{3} = 0$$

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

$$r + s = -\frac{2}{3} rs = -\frac{5}{3}$$

Two numbers r and s sum up to -\frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{2}{3} = -\frac{1}{3}. 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{1}{3} - u s = -\frac{1}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{3}

$$(-\frac{1}{3} - u) (-\frac{1}{3} + u) = -\frac{5}{3}$$

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

$$\frac{1}{9} - u^2 = -\frac{5}{3}$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = -\frac{5}{3}-\frac{1}{9} = -\frac{16}{9}$$

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

$$u^2 = \frac{16}{9} u = \pm\sqrt{\frac{16}{9}} = \pm \frac{4}{3}$$

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

$$r =-\frac{1}{3} - \frac{4}{3} = -1.667 s = -\frac{1}{3} + \frac{4}{3} = 1.000$$

Evaluate

$$\left(1-x\right)\left(3x+5\right)$$