Solve -4x^2+260x+1400

$$-4 { x }^{ 2 } +260x+1400$$

Factor

$$-4\left(x-70\right)\left(x+5\right)$$

Solution Steps

Factor out 4.

$$4\left(-x^{2}+65x+350\right)$$

Consider -x^{2}+65x+350. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+350. To find a and b, set up a system to be solved.

$$a+b=65 ab=-350=-350$$

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 -350.

$$-1,350 -2,175 -5,70 -7,50 -10,35 -14,25$$

Calculate the sum for each pair.

$$-1+350=349 -2+175=173 -5+70=65 -7+50=43 -10+35=25 -14+25=11$$

The solution is the pair that gives sum 65.

$$a=70 b=-5$$

Rewrite -x^{2}+65x+350 as \left(-x^{2}+70x\right)+\left(-5x+350\right).

$$\left(-x^{2}+70x\right)+\left(-5x+350\right)$$

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

$$-x\left(x-70\right)-5\left(x-70\right)$$

Factor out common term x-70 by using distributive property.

$$\left(x-70\right)\left(-x-5\right)$$

Rewrite the complete factored expression.

$$4\left(x-70\right)\left(-x-5\right)$$

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.

$$-4x^{2}+260x+1400=0$$

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.

$$x=\frac{-260±\sqrt{260^{2}-4\left(-4\right)\times 1400}}{2\left(-4\right)}$$

Square 260.

$$x=\frac{-260±\sqrt{67600-4\left(-4\right)\times 1400}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-260±\sqrt{67600+16\times 1400}}{2\left(-4\right)}$$

Multiply 16 times 1400.

$$x=\frac{-260±\sqrt{67600+22400}}{2\left(-4\right)}$$

Add 67600 to 22400.

$$x=\frac{-260±\sqrt{90000}}{2\left(-4\right)}$$

Take the square root of 90000.

$$x=\frac{-260±300}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-260±300}{-8}$$

Now solve the equation x=\frac{-260±300}{-8} when ± is plus. Add -260 to 300.

$$x=\frac{40}{-8}$$

Divide 40 by -8.

$$x=-5$$

Now solve the equation x=\frac{-260±300}{-8} when ± is minus. Subtract 300 from -260.

$$x=-\frac{560}{-8}$$

Divide -560 by -8.

$$x=70$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -5 for x_{1} and 70 for x_{2}.

$$-4x^{2}+260x+1400=-4\left(x-\left(-5\right)\right)\left(x-70\right)$$

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

$$-4x^{2}+260x+1400=-4\left(x+5\right)\left(x-70\right)$$

Evaluate

$$-4\left(x-70\right)\left(x+5\right)$$