Problem
Quadratic expression problems often require three connected skills: factoring a quadratic polynomial, simplifying an algebraic expression using that factorization, and solving for the zeros of the quadratic. Use the quadratic expression \(x^2 + 5\cdot x + 6\) for all parts below.
Part A: Factor the quadratic expression
A common target form is \((x+p)\cdot(x+q)\), because expanding gives \[ (x+p)\cdot(x+q) = x^2 + (p+q)\cdot x + p\cdot q. \] Matching coefficients with \(x^2 + 5\cdot x + 6\) requires: \[ p+q = 5,\quad p\cdot q = 6. \]
- Find two numbers with product \(6\) and sum \(5\): \(2\) and \(3\).
- Substitute \(p=2\), \(q=3\): \[ x^2 + 5\cdot x + 6 = (x+2)\cdot(x+3). \]
Factored form: \(\;x^2 + 5\cdot x + 6 = (x+2)\cdot(x+3)\)
Part B: Simplify \(\frac{x^2 + 5\cdot x + 6}{x+2}\)
Substitute the factorization from Part A: \[ \frac{x^2 + 5\cdot x + 6}{x+2} = \frac{(x+2)\cdot(x+3)}{x+2}. \]
- Cancel the common factor \(x+2\) (only allowed when \(x+2\ne 0\)): \[ \frac{(x+2)\cdot(x+3)}{x+2} = x+3. \]
- Record the restriction created by the original denominator: \[ x \ne -2. \]
Simplified result: \(\;\frac{x^2 + 5\cdot x + 6}{x+2} = x+3\;\) for \(\;x\ne -2\)
The simplified expression is equivalent to the original everywhere the original is defined; \(x=-2\) remains excluded.
Part C: Solve \(x^2 + 5\cdot x + 6 = 0\)
Use the factored form: \[ (x+2)\cdot(x+3) = 0. \] A product equals zero exactly when at least one factor equals zero.
- Set \(x+2=0\) to get \(x=-2\).
- Set \(x+3=0\) to get \(x=-3\).
Solutions (zeros/roots): \(\;x=-2\) and \(\;x=-3\)
Summary table for these quadratic expression problems
| Task | Result | Notes |
|---|---|---|
| Factor \(x^2 + 5\cdot x + 6\) | \((x+2)\cdot(x+3)\) | Choose numbers \(2\) and \(3\) so sum \(5\), product \(6\). |
| Simplify \(\frac{x^2 + 5\cdot x + 6}{x+2}\) | \(x+3\) | Valid only for \(\;x\ne -2\) due to the original denominator. |
| Solve \(x^2 + 5\cdot x + 6 = 0\) | \(x=-2,\;x=-3\) | Zero-product property after factoring. |
Visualization: area model for \((x+2)\cdot(x+3)=x^2+5\cdot x+6\)
Common pitfalls
- Sign and pairing errors: for \(x^2 + b\cdot x + c\), the chosen numbers must add to \(b\) and multiply to \(c\).
- Invalid cancellation: canceling \((x+2)\) requires \(x\ne -2\); domain restrictions must be preserved.
- Mixing tasks: simplifying \(\frac{(x+2)\cdot(x+3)}{x+2}\) does not “remove” the excluded value \(x=-2\) from the original expression.