Solve 3f^2+15f-42

$$3 { f }^{ 2 } +15f-42$$

Factor

$$3\left(f-2\right)\left(f+7\right)$$

Solution Steps

Factor out 3.

$$3\left(f^{2}+5f-14\right)$$

Consider f^{2}+5f-14. Factor the expression by grouping. First, the expression needs to be rewritten as f^{2}+af+bf-14. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-14\right)=-14$$

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 -14.

$$-1,14 -2,7$$

Calculate the sum for each pair.

$$-1+14=13 -2+7=5$$

The solution is the pair that gives sum 5.

$$a=-2 b=7$$

Rewrite f^{2}+5f-14 as \left(f^{2}-2f\right)+\left(7f-14\right).

$$\left(f^{2}-2f\right)+\left(7f-14\right)$$

Factor out f in the first and 7 in the second group.

$$f\left(f-2\right)+7\left(f-2\right)$$

Factor out common term f-2 by using distributive property.

$$\left(f-2\right)\left(f+7\right)$$

Rewrite the complete factored expression.

$$3\left(f-2\right)\left(f+7\right)$$

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.

$$3f^{2}+15f-42=0$$

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.

$$f=\frac{-15±\sqrt{15^{2}-4\times 3\left(-42\right)}}{2\times 3}$$

Square 15.

$$f=\frac{-15±\sqrt{225-4\times 3\left(-42\right)}}{2\times 3}$$

Multiply -4 times 3.

$$f=\frac{-15±\sqrt{225-12\left(-42\right)}}{2\times 3}$$

Multiply -12 times -42.

$$f=\frac{-15±\sqrt{225+504}}{2\times 3}$$

Add 225 to 504.

$$f=\frac{-15±\sqrt{729}}{2\times 3}$$

Take the square root of 729.

$$f=\frac{-15±27}{2\times 3}$$

Multiply 2 times 3.

$$f=\frac{-15±27}{6}$$

Now solve the equation f=\frac{-15±27}{6} when ± is plus. Add -15 to 27.

$$f=\frac{12}{6}$$

Divide 12 by 6.

$$f=2$$

Now solve the equation f=\frac{-15±27}{6} when ± is minus. Subtract 27 from -15.

$$f=-\frac{42}{6}$$

Divide -42 by 6.

$$f=-7$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 2 for x_{1} and -7 for x_{2}.

$$3f^{2}+15f-42=3\left(f-2\right)\left(f-\left(-7\right)\right)$$

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

$$3f^{2}+15f-42=3\left(f-2\right)\left(f+7\right)$$

Evaluate

$$3\left(f-2\right)\left(f+7\right)$$