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Solving polynomials with the unknown "b" from \(ax^2 + bx + c\)

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Finding the Unknown b in x² + bx + c

When a quadratic trinomial x squared plus bx plus c factors as (x+p)(x+q), the coefficient b equals the sum of p and q. Learn to find the missing b from a factored form, from known factor pairs of c, and how it relates to finding the missing c.

Finding the unknown b in x² + bx + c

A quadratic trinomial x² + bx + c factors as (x + p)(x + q) when p and q multiply to give c and add to give b. If you're given c along with the two factors, or the factored form directly, finding b just means adding the two numbers.

Finding b in x²+bx+c Given the factored form (x+3)(x+6), the coefficient b in x squared plus bx plus 18 equals the sum of 3 and 6, which is 9. Factored form: (x + 3)(x + 6) c = product of constants:c = 3 × 6 = 18 b = sum of constants:b = 3 + 6 = 9 x² + 9x + 18 = (x + 3)(x + 6)
From the factored form (x+3)(x+6), b is the sum of the constants: 3 + 6 = 9.

Worked example

Given the factored form (x + 3)(x + 6), find b in x² + bx + 18. Add the two numbers inside the parentheses: b = 3 + 6 = 9. (Multiplying them instead gives c: 3 × 6 = 18, confirming the trinomial.) So x² + 9x + 18 = (x + 3)(x + 6).

Finding factors when only c is known

If you're only given c, list pairs of numbers that multiply to c, then add each pair to see which gives a matching b. For c = 18, the pairs are (1,18), (2,9), and (3,6) — each pair adds to a different possible b.

How this connects

This is the mirror skill to finding the unknown c: both come from the same pair of numbers, just combined differently (sum for b, product for c). Together they're the foundation of factoring trinomials in general.

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