Cubic with a Factor and a Remainder: Recovering Coefficients
During a revision session, a candidate is given the cubic p(x)=x³+ax²+bx+6, where a and b are real constants. The examiner states that (x−2) is a factor of p(x), and that when p(x) is divided by (x−1) the remainder is 4. The candidate substitutes x=2 and x=1, obtaining two equations, but then hesitates because the second condition is a remainder rather than a factor and the signs feel reversed. A classmate claims a=4 and b=−7 after an arithmetic slip. The candidate must find a and b, factorise p(x) completely, solve p(x)≤0, and decide whether (x−4) can be a factor without performing long division. A calculator is allowed, but full method marks require a clear statement of the factor theorem and a sign diagram. The candidate has ten minutes and must present a solution that can be checked by substitution at each stage. The examiner is particularly careful to award method marks for the correct use of p(2)=0 and p(1)=4.
1. Find the values of a and b. Show the equations obtained from the factor theorem and remainder theorem, and check your values.
Read the worked answer
Since (x−2) is a factor, the factor theorem gives p(2)=0:
8+4a+2b+6=0 ⇒ 4a+2b+14=0 ⇒ 2a+b=−7.
Since the remainder on division by (x−1) is 4, p(1)=4:
1+a+b+6=4 ⇒ a+b+7=4 ⇒ a+b=−3.
Subtract the second equation from the first: (2a+b)−(a+b)=−7−(−3), so a=−4. Then b=−3−a=1.
Check: p(2)=8−16+2+6=0 and p(1)=1−4+1+6=4. Both conditions hold, so a=−4 and b=1.
2. Factorise p(x) completely and solve p(x)≤0. Use a sign diagram or an equivalent argument.
Read the worked answer
With a=−4 and b=1, p(x)=x³−4x²+x+6. Dividing by the known factor (x−2) gives x²−2x−3, which factorises as (x−3)(x+1). Hence p(x)=(x−2)(x−3)(x+1).
The critical values are x=−1, 2, 3. The leading coefficient is positive, so the cubic is positive for large positive x. Checking one value in each interval gives:
- x<−1: p(−2)=−20<0,
- −1<x<2: p(0)=6>0,
- 2<x<3: p(2.5)=−0.875<0,
- x>3: p(4)=10>0.
Therefore p(x)≤0 for x≤−1 or 2≤x≤3, i.e. (−∞,−1]∪[2,3]. The inequality is non-strict, so the roots themselves are included.
3. Without long division, find the remainder when p(x) is divided by (x−4) and decide whether (x−4) is a factor. Explain what would have to be true for it to be a factor.
Read the worked answer
By the remainder theorem, the remainder on division by (x−4) is p(4):
p(4)=4³−4(4²)+4+6=64−64+4+6=10.
Since the remainder is 10, not 0, (x−4) is not a factor. The factor theorem states that (x−4) would be a factor exactly when p(4)=0; here that would require 10=0, which is false. No long division is needed because substitution supplies both the remainder and the factor decision.