Solve 3x^2-2x+4text{247}2x^2+3x-5

$$3 x ^ { 2 } - 2 x + 4 \text { 247 } 2 x ^ { 2 } + 3 x - 5$$

Evaluate

$$1979x^{2}+x-5$$

Solution Steps

Multiply 4 and 247 to get 988.

$$3x^{2}-2x+988\times 2x^{2}+3x-5$$

Multiply 988 and 2 to get 1976.

$$3x^{2}-2x+1976x^{2}+3x-5$$

Combine 3x^{2} and 1976x^{2} to get 1979x^{2}.

$$1979x^{2}-2x+3x-5$$

Combine -2x and 3x to get x.

$$1979x^{2}+x-5$$

Factor

$$1979\left(x-\frac{-\sqrt{39581}-1}{3958}\right)\left(x-\frac{\sqrt{39581}-1}{3958}\right)$$

Steps Using the Quadratic Formula

Multiply 4 and 247 to get 988.

$$factor(3x^{2}-2x+988\times 2x^{2}+3x-5)$$

Multiply 988 and 2 to get 1976.

$$factor(3x^{2}-2x+1976x^{2}+3x-5)$$

Combine 3x^{2} and 1976x^{2} to get 1979x^{2}.

$$factor(1979x^{2}-2x+3x-5)$$

Combine -2x and 3x to get x.

$$factor(1979x^{2}+x-5)$$

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.

$$1979x^{2}+x-5=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1979\left(-5\right)}}{2\times 1979}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 1979\left(-5\right)}}{2\times 1979}$$

Multiply -4 times 1979.

$$x=\frac{-1±\sqrt{1-7916\left(-5\right)}}{2\times 1979}$$

Multiply -7916 times -5.

$$x=\frac{-1±\sqrt{1+39580}}{2\times 1979}$$

Add 1 to 39580.

$$x=\frac{-1±\sqrt{39581}}{2\times 1979}$$

Multiply 2 times 1979.

$$x=\frac{-1±\sqrt{39581}}{3958}$$

Now solve the equation x=\frac{-1±\sqrt{39581}}{3958} when ± is plus. Add -1 to \sqrt{39581}.

$$x=\frac{\sqrt{39581}-1}{3958}$$

Now solve the equation x=\frac{-1±\sqrt{39581}}{3958} when ± is minus. Subtract \sqrt{39581} from -1.

$$x=\frac{-\sqrt{39581}-1}{3958}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-1+\sqrt{39581}}{3958} for x_{1} and \frac{-1-\sqrt{39581}}{3958} for x_{2}.

$$1979x^{2}+x-5=1979\left(x-\frac{\sqrt{39581}-1}{3958}\right)\left(x-\frac{-\sqrt{39581}-1}{3958}\right)$$