Solve p(x)=4-3x-x^2

$$p ( x ) = 4 - 3 x - x ^ { 2 }$$

Factor

$$\left(1-x\right)\left(x+4\right)$$

Solution Steps

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

$$-x^{2}-3x+4$$

Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-4=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=1 b=-4$$

Rewrite -x^{2}-3x+4 as \left(-x^{2}+x\right)+\left(-4x+4\right).

$$\left(-x^{2}+x\right)+\left(-4x+4\right)$$

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

$$x\left(-x+1\right)+4\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(x+4\right)$$

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.

$$-x^{2}-3x+4=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2\left(-1\right)}$$

Add 9 to 16.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

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

The opposite of -3 is 3.

$$x=\frac{3±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±5}{-2}$$

Now solve the equation x=\frac{3±5}{-2} when ± is plus. Add 3 to 5.

$$x=\frac{8}{-2}$$

Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{3±5}{-2} when ± is minus. Subtract 5 from 3.

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

Divide -2 by -2.

$$x=1$$

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

$$-x^{2}-3x+4=-\left(x-\left(-4\right)\right)\left(x-1\right)$$

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

$$-x^{2}-3x+4=-\left(x+4\right)\left(x-1\right)$$

Evaluate

$$\left(1-x\right)\left(x+4\right)$$