What you’ll learn
Factor ax² + bx + c.
Key idea
Unequal leading coefficients make the placement of constants matter; compare the cross-product sum.
Unequal leading coefficients make constant placement matter; verify the cross coefficient as well as first and last products.
Worked example
Enter the exact value or ordered values in the format requested below.
Factor 2x²-9x+10 into integer binomials, preserving leading coefficient. Enter product of factors.
- Leading product is 2x² and constant product -5×-2=10; cross coefficient 2×(-2)+(-5)=-9.
- (2x-5)(x-2) satisfies all coefficients. Swapping constants gives cross coefficient -12, so (2x-2)(x-5) fails the middle term.
Answer: (2x-5)(x-2)
Quick Check
Evaluate a worked attempt at factor ax² + bx + c, identify the first invalid step, and provide the correction.
Factor 2x²+11x+12: Step 1 chooses leading factors 2x and x, constant candidates 3,4. Step 2 accepts (2x+4)(x+3) solely because leading and constant products match. Locate the first invalid step and repair it.