Factor theorem and factoring a cubic completely
Use the factor theorem to test a possible root, then factor a cubic completely. Unlimited questions, five difficulty levels, and a full worked solution every time — free.
Try one
P(1) = 0. Use the factor theorem to factor P(x) completely.
P(x) = x3 - 3x2 - x + 3
- (x + 1)(x - 1)(x + 3)
- (x + 1)(x + 1)(x - 3)
- (x - 1)(x - 1)(x - 3)
- (x + 1)(x - 1)(x - 3)
Answer: (x + 1)(x - 1)(x - 3)
See one solved, step by step
P(-4) = 0. Use the factor theorem to factor P(x) completely.
P(x) = 3x3 - 6x2 - 48x + 96
- (x + 4)(x - 2)(x - 4)
- 3(x + 4)(x - 2)(x - 4)
- 3(x - 4)(x - 2)(x - 4)
- 3(x + 4)(x + 2)(x - 4)
📘 Worked solution
1P(−4) = 0, so by the factor theorem (x + 4) is a factor of P(x).The factor theorem: if P(k) = 0, then (x - k) is a factor.
2Dividing P(x) by (x + 4) leaves a quotient that factors further.Polynomial division (or synthetic division) reduces the cubic to a quadratic you can factor directly.
3P(x) = 3(x + 4)(x - 2)(x - 4)The three roots of P(x) = 0 are -4, 2, 4, one factor per root, with leading coefficient 3.
Ready to master it?
Jump into unlimited practice — your Winny Score tracks you to 100. And if it won't click, a real tutor will make it click: first lesson free.
Start practicing