Question 1
The polynomial p is such that where a and b are constants.
It is given that:
- x+2 is a factor of p(x)
- when p(x) is divided by x-3 the remainder is 40.
Find the values of a and b.
The polynomial p is such that p(x)=x3+ax2+bx−2, where a and b are constants.
It is given that:
- x+2 is a factor of p(x)
- when p(x) is divided by x-3 the remainder is 40.
Find the values of a and b.
-8+4a-2b-2=0 oe
27+9a+3b-2=40 oe
Solves their linear equations in a and b to find one unknown.
a=2 and b=-1 nfww
A1 for a=2 or b=-1.
The polynomial p is such that p(x)=3x3−7x2+ax+b, where a and b are integers.
It is given that p′(−1)=21 and that x-2 is a factor of p(x).
Find the values of a and b.
p′(x)=9x2−14x+a soi
p′(−1)=9+14+a=21a=−2
p(2)=24-28+2 a+b=0 soi
Each term must be simplified, allow using their a
b=8
Hence write p(x) as a product of linear factors with integer coefficients.
[x−2](3x2−x−4) soi
M1 for quadratic factor with two terms correct
A1 must be from correct a and b
(x-2)(x+1)(3 x-4)
A1
A1 must be from correct a and b
DO NOT USE A CALCULATOR IN THIS QUESTION.
The polynomial p is defined by p(x)=ax3−3x2−3x+b, where a and b are constants.
Given that x=2 and x=-1 are roots of the equation p(x)=0, find a and b.
8 a-12-6+b=0 oe
and -a-3+3+b=0 oe
and a=2, b=2
B1 for 8 a-12-6+b=0 oe
B1 for -a-3+3+b=0 oe
Solve the equation p(x)=0.
and x=2,21,−1
OR
[2x3−3x2−3x+2=](x+1),(2x2−5x+2) or (x−2),(2x2+x−1)
and correct factorisation or method of solution of the quadratic
and x=2,21,−1
OR
2−1+x=−(2−3) or 2−1+x=23
or for 2×−1×x=2−2 or 2×−1×x=−1 and x=2,21,−1
B1 for
[2x3−3x2−3x+2=]
(x-2) and (x+1) seen
or ( x2−x−2 ) seen
or (x+1),(2x2−5x+2)
or (x−2),(2x2+x−1)
B1 for 2−1+x=−(2−3) or 23
or for 2×−1×x=2−2 or -1