Question
The polynomial \(P(x) = {x^3} + a{x^2} + bx + 2\) is divisible by (x +1) and by (x − 2) .
Find the value of a and of b, where \(a,{\text{ }}b \in \mathbb{R}\) .
▶️Answer/Explanation
Markscheme
METHOD 1
As (x +1) is a factor of P(x), then P(−1) = 0 (M1)
\( \Rightarrow a – b + 1 = 0\) (or equivalent) A1
As (x − 2) is a factor of P(x), then P(2) = 0 (M1)
\( \Rightarrow 4a + 2b + 10 = 0\) (or equivalent) A1
Attempting to solve for a and b M1
a = −2 and b = −1 A1 N1
[6 marks]
METHOD 2
By inspection third factor must be x −1. (M1)A1
\((x + 1)(x – 2)(x – 1) = {x^3} – 2{x^2} – x + 2\) (M1)A1
Equating coefficients a = −2, b = −1 (M1)A1 N1
[6 marks]
METHOD 3
Considering \(\frac{{P(x)}}{{{x^2} – x – 2}}\) or equivalent (M1)
\(\frac{{P(x)}}{{{x^2} – x – 2}} = (x + a + 1) + \frac{{(a + b + 3)x + 2(a + 2)}}{{{x^2} – x – 2}}\) A1A1
Recognising that \((a + b + 3)x + 2(a + 2) = 0\) (M1)
Attempting to solve for a and b M1
a = −2 and b = −1 A1 N1
[6 marks]
Question
When \(f(x) = {x^4} + 3{x^3} + p{x^2} – 2x + q\) is divided by (x − 2) the remainder is 15, and (x + 3) is a factor of f(x) .
Find the values of p and q .
▶️Answer/Explanation
Markscheme
\(f(2) = 16 + 24 + 4p – 4 + q = 15\) M1
\( \Rightarrow 4p + q = – 21\) A1
\(f( – 3) = 81 – 81 + 9p + 6 + q = 0\) M1
\( \Rightarrow 9p + q = – 6\) A1
\( \Rightarrow p = 3{\text{ and }}q = – 33\) A1A1 N0
[6 marks]
Question
When the function \(q(x) = {x^3} + k{x^2} – 7x + 3\) is divided by (x + 1) the remainder is seven times the remainder that is found when the function is divided by (x + 2) .
Find the value of k .
▶️Answer/Explanation
Markscheme
\(q( – 1) = k + 9\) M1A1
\(q( – 2) = 4k + 9\) A1
\(k + 9 = 7(4k + 9)\) M1
\(k = – 2\) A1
Notes: The first M1 is for one substitution and the consequent equations.
Accept expressions for \(q( – 1)\) and \(q( – 2)\) that are not simplified.
[5 marks]
Question
When \(3{x^5} – ax + b\) is divided by x −1 and x +1 the remainders are equal. Given that a , \(b \in \mathbb{R}\) , find
(a) the value of a ;
(b) the set of values of b .
▶️Answer/Explanation
Markscheme
(a) \(f(1) = 3 – a + b\) (A1)
\(f( – 1) = – 3 + a + b\) (A1)
\(3 – a + b = – 3 + a + b\) M1
\(2a = 6\)
\(a = 3\) A1 N4
(b) b is any real number A1
[5 marks]