Question
A polynomial \(p(x)\) with real coefficients is of degree five. The equation \(p(x) = 0\) has a complex root 2 + i. The graph of \(y = p(x)\) has the x-axis as a tangent at (2, 0) and intersects the coordinate axes at (−1, 0) and (0, 4).
Find \(p(x)\) in factorised form with real coefficients.
▶️Answer/Explanation
Markscheme
other root is 2 – i (A1)
a quadratic factor is therefore \((x – 2 + i)(x – 2 – i)\) (M1)
\( = {x^2} – 4x + 5\) A1
x + 1 is a factor A1
\({(x – 2)^2}\) is a factor A1
\(p(x) = a(x + 1){(x – 2)^2}({x^2} – 4x + 5)\) (M1)
\(p(0) = 4 \Rightarrow a = \frac{1}{5}\) A1
\(p(x) = \frac{1}{5}(x + 1){(x – 2)^2}({x^2} – 4x + 5)\)
[7 marks]
Examiners report
Whilst most candidates knew that another root was \(2 – {\text{i}}\) , much fewer were able to find the quadratic factor. Surprisingly few candidates knew that \(\left( {x – 2} \right)\) must be a repeated factor and less surprisingly many did not recognise that the whole expression needed to be multiplied by \(\frac{1}{5}\).
Question
When \({x^2} + 4x – b\) is divided by \(x – a\) the remainder is 2.
Given that \(a,{\text{ }}b \in \mathbb{R}\), find the smallest possible value for \(b\).
▶️Answer/Explanation
Markscheme
\({a^2} + 4a – b = 2\) M1A1
EITHER
\({a^2} + 4a – (b + 2) = 0\)
as \(a\) is real \( \Rightarrow 16 + 4(b + 2) \geqslant 0\) M1A1
OR
\(b = {a^2} + 4a – 2\) M1
\( = {(a + 2)^2} – 6\) (A1)
THEN
\(b \geqslant – 6\)
hence smallest possible value for \(b\) is \( – 6\) A1
[5 marks]
Examiners report
For quite a difficult question, there were many good solutions for this, including many different methods. It was disturbing to see how many students did not seem to be aware of the remainder theorem, instead choosing to divide the polynomial.
Question
The polynomial \({x^4} + p{x^3} + q{x^2} + rx + 6\) is exactly divisible by each of \(\left( {x – 1} \right)\), \(\left( {x – 2} \right)\) and \(\left( {x – 3} \right)\).
Find the values of \(p\), \(q\) and \(r\).
▶️Answer/Explanation
Markscheme
METHOD 1
substitute each of \(x\) = 1,2 and 3 into the quartic and equate to zero (M1)
\(p + q + r = – 7\)
\(4p + 2q + r = – 11\) or equivalent (A2)
\(9p + 3q + r = – 29\)
Note: Award A2 for all three equations correct, A1 for two correct.
attempting to solve the system of equations (M1)
\(p\) = −7, \(q\) = 17, \(r\) = −17 A1
Note: Only award M1 when some numerical values are found when solving algebraically or using GDC.
METHOD 2
attempt to find fourth factor (M1)
\(\left( {x – 1} \right)\) A1
attempt to expand \({\left( {x – 1} \right)^2}\left( {x – 2} \right)\left( {x – 3} \right)\) M1
\({x^4} – 7{x^3} + 17{x^2} – 17x + 6\) (\(p\) = −7, \(q\) = 17, \(r\) = −17) A2
Note: Award A2 for all three values correct, A1 for two correct.
Note: Accept long / synthetic division.
[5 marks]
Examiners report
Question
Consider the cubic function f(x) = ax3 + bx2 + 3x +4. Find the values of a and b in each of the following cases
(a) f(x) is divisible by (x – 1) and leaves a remainder 6 when divided by (x + 1).
(b) f(x) is divisible by (x2 – 1).
(c) f(x) leaves a remainder -3x + 3 when divided by (x2 – 1).
▶️Answer/Explanation
Ans
(a) a = -6 b = -1 (b) a = -3, b = -4 (c) a = -6 b = -1
Question
Consider the cubic function f (x) = 2x3 + ax2 + bx + c. Find the values of a,b,c
(a) if the graph of the function passes through the points (1,0), (–1 ,2), and (0,3).
(b) if the graph of the function passes through the points (1,0), (–1,0), and (3,0).
▶️Answer/Explanation
Ans
(a) a = -2 b = -3 c = 3 (b) a = -6, b = -2, c = 6
Question
Consider the polynomial \(f(x) =4x^3-4x^2-5x+3.\)
(a) Show that x = –1 is a root and find the other two roots by using long division.
(b) Express the polynomial in the form f(x) = (x – a)(bx – c)(dx – e), where a,b,c,d,e∈Z.
▶️Answer/Explanation
Ans
(a) \(x = -1, x=\frac{1}{2}, x = \frac{3}{2}\) (b) f(x) = (x+1)(2x-1)(2x-3)