4Remainder by a quadraticP(x)÷(x−a)(x−b) leaves px+qpa+q=P(a),pb+q=P(b)
5Cubic: sum and productr1+r2+r3=−ab,r1r2r3=−adr1r2+r2r3+r3r1=ac;r1,r2,r3 the roots of ax3+bx2+cx+d=0
6Sum of reciprocalsr11+r21+r31=−dcr1,r2,r3 roots of ax3+bx2+cx+d=0
7Factored formP(k)=a(k−r1)(k−r2)(k−r3)r1,r2,r3 the roots of P,a its leading coefficient
8Roots in APIf a cubic's roots are in AP, write them m−d, m, m+d: the sum of roots gives m at once, and m is itself a root.
9Values on a lineIf a cubic P has P(1),P(2),P(3) on a line mx+c, then P(x)−(mx+c) is 0 at 1, 2 and 3, so it is (x−1)(x−2)(x−3) times P's leading coefficient.
10Surd inputTo evaluate P(x) at x=a+b, use (x−a)2=b to cut the powers down before substituting.
11Compare coefficientsWhen two polynomials are equal for every x, match the coefficients of each power: x3 with x3, x2 with x2, down to the constants.
Test yourself
10 questions · 3 easy · 4 medium · 3 hard
Tap an option to answer. The first tap counts.
Q1Easy
Long division of P(x) by x−3 gives the quotient x2+2x−1 and the remainder 5. Find P(2).
Correct answer·−2
Why
Write P(x)=(x−3)(x2+2x−1)+5 and substitute directly.
At x=2: (2−3)(4+4−1)+5=−7+5.
So P(2)=−2.
Q2Easy
Asked in IPMAT Indore 2023
A polynomial p(x) leaves a remainder 2 when divided by (x−1) and a remainder 1 when divided by (x−2). The remainder when p(x) is divided by (x−1)(x−2) is
Correct answer·3−x
Why
A quadratic divisor can only leave a remainder of degree 1 or less, say px+q.
The two given remainders are its values: p+q=2 at x=1 and 2p+q=1 at x=2, so p=−1 and q=3.
That makes the remainder 3−x.
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