randRangeNonZero( -10, 10 ) randRangeNonZero( -10, 10 )
1 SQUARE*A*B A*B SQUARE*(-A-B) -A-B

Factor the expression below completely. All coefficients should be integers.

`plus(SQUARE + "x^2") + plus( LINEAR + "x" ) + CONSTANT`

(x-A)(x-B)

When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together:

`\qquad \begin{eqnarray} (x + a)(x + b) \quad&=&\quad xx &+& xb + ax &+& ab \\ \\ &=&\quad x^2 &+& \green{(a + b)}x &+& \blue{ab} \end{eqnarray}`

`\qquad \begin{eqnarray} \hphantom{(x + a)(x + b) \quad}&\hphantom{=}&\hphantom{\quad xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=&\quad x^2 & SIMPLELINEAR >= 0 ? "+" : ""& plus( "\\green{" + SIMPLELINEAR + "}x" )& SIMPLECONSTANT >= 0 ? "+" : ""& plus( "\\blue{" + SIMPLECONSTANT + "}" ) \end{eqnarray}`

The coefficient on the `x` term is `SIMPLELINEAR` and the constant term is `SIMPLECONSTANT`, so to reverse the steps above, we need to find two numbers that add up to `SIMPLELINEAR` and multiply to `SIMPLECONSTANT`.

You can try out different factors of `\blue{SIMPLECONSTANT}` to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for `\pink{a}` and `\pink{b}`:

`\qquad \pink{a} + \pink{b} = \green{SIMPLELINEAR}`

`\qquad \pink{a} \times \pink{b} = \blue{SIMPLECONSTANT}`

The two numbers `\pink{-A}` and `\pink{-B}` satisfy both conditions:

`\qquad \pink{-A} + \pink{-B} = \green{SIMPLELINEAR}`

`\qquad \pink{-A} \times \pink{-B} = \blue{SIMPLECONSTANT}`

So we can factor the expression as: `(x A < 0 ? "+" : "" \pink{-A})(x B < 0 ? "+" : "" \pink{-B})`