AMC 10 · 2002 · #10
Grade 8 arithmeticPick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Instead of multiplying everything out, look at the shape of the left side: it is (2x+3) times one thing plus (2x+3) times another thing. Tool #16 (Change Focus) says read that structure rather than expanding, because the shared factor (2x+3) can be pulled out. Once it is factored into a product that equals zero, tool #7 (Identify Subproblems) takes over: a product is zero exactly when one of its factors is zero, so the single hard equation splits into two easy linear equations. Solve each for one root, then add the roots.
Pull out the shared factor
Both terms carry the factor (2x+3), so pull it out: (2x+3)[(x-4)+(x-6)] = (2x+3)(2x-10) = 0.
When every piece carries the same factor, you can grab that factor once instead of multiplying it back in.
6.EE.A.3Change Focus Count The ComplementSplit into two linear equations
A product is zero only if a factor is zero, so 2x+3=0 gives x=-3/2 and 2x-10=0 gives x=5.
If two numbers multiply to zero, one of them had to be zero, so each factor hands you one root.
If two numbers multiply to zero, one of them had to be zero, so each factor hands you one root.
▸ Why?
Two nonzero numbers can never multiply to zero, so no root can hide outside the factors.
▸ Why?
Pulling the shared factor out front is what turned the sum into a product in the first place.
Add the two roots
Add the roots, writing 5 as 10/2: -3/2+10/2 = 7/2, which is choice (A).
Adding a negative fraction to a whole number is just lining them up over the same denominator.
7.NS.A.1Identify SubproblemsWhen two terms share a factor, pull it out first — a product that equals zero breaks the problem into one easy equation per factor.
- Pull out the shared factor
- Split into two linear equations
- Add the two roots