top of page

Factors or solutions where the x² coefficient = 1

CONCEPTS

Terms

• Standard form of a polynomial equation: ax² + bx + c = 0

• Factored form: (x + a)(x + b) = 0

• Factors: (x + a) & (x + b)

• Solutions, Roots, Zeros: x = -a, x = -b

• Zeros: the value of the factor when it crosses the x axis

• Binomial: a polynomial with only 2 terms (Examples: x² + x, x² + y², x² + 2)


Tips

• It's sometimes easier to go backwards and see what answers work vs solving the problem

To factor a quadratic and determine the factors, use either the method under “Process for Simple Factoring”, or use the quadratic formula. Sometimes it’s easier to use the quadratic formula when there’s a leading coefficient > 1 (ex: 2x² + 3x + 4)

• Always put the solution, root or zero back into the original equation to confirm result = 0.

PRACTICE PROBLEMS

Question 1 What are the solutions for the polynomial x⁴ + 5x² + 2 = 0


-- Answer -- x = √ ((-5 +- √17)/2) AND √(- (-5 +- √17)/2)

let y = x²

y² + 5y + 2 = 0

Use the quadratic equation to find the solutions:

y1, y2 = ((-5) +- √(5² - 4*1*2)) / (2 * 1) = (-5 +- √17)/2

y = x² => x² = (-5 +- √17)/2 AND x² = - (-5 +- √17)/2

x = √ ((-5 +- √17)/2) AND √(- (-5 +- √17)/2)

ACT PRACTICE PROBLEMS

bottom of page