Solve 30+7x-x^2

$$30 + 7 x - x ^ { 2 }$$

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+7x+30$$

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=7 ab=-30=-30$$

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.

$$a=10 b=-3$$

Rewrite -x^{2}+7x+30 as \left(-x^{2}+10x\right)+\left(-3x+30\right).

$$\left(-x^{2}+10x\right)+\left(-3x+30\right)$$

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

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

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

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

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.

$$-x^{2}+7x+30=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-1\right)\times 30}}{2\left(-1\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-1\right)\times 30}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-7±\sqrt{49+4\times 30}}{2\left(-1\right)}$$

Multiply 4 times 30.

$$x=\frac{-7±\sqrt{49+120}}{2\left(-1\right)}$$

Add 49 to 120.

$$x=\frac{-7±\sqrt{169}}{2\left(-1\right)}$$

Take the square root of 169.

$$x=\frac{-7±13}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-7±13}{-2}$$

Now solve the equation x=\frac{-7±13}{-2} when ± is plus. Add -7 to 13.

$$x=\frac{6}{-2}$$

Divide 6 by -2.

$$x=-3$$

Now solve the equation x=\frac{-7±13}{-2} when ± is minus. Subtract 13 from -7.

$$x=-\frac{20}{-2}$$

Divide -20 by -2.

$$x=10$$

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

$$-x^{2}+7x+30=-\left(x-\left(-3\right)\right)\left(x-10\right)$$

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

$$-x^{2}+7x+30=-\left(x+3\right)\left(x-10\right)$$

Evaluate

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