polynominal b6 2009 alevel, Quizzes of Mathematics

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B6 Polynomials

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Time: 377 minutes Marks: 308 marks

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Q1. Find f( x ). (Total 4 marks) Q2. p( x ) = 30 x^3 − 7 x^2 − 7 x + 2 (a) Prove that (2 x + 1) is a factor of p( x ) (2) (b) Factorise p( x ) completely. (3) (c) Prove that there are no real solutions to the equation (5) (Total 10 marks) Q3. p( x ) = x^3 − 5 x^2 + 3 x + a , where a is a constant. Given that x − 3 is a factor of p( x ), find the value of a Circle your answer. −9 −3 3 9 (Total 1 mark) Q4. p( x ) = 2 x^3 + 7 x^2 + 2 x − 3 (a) Use the factor theorem to prove that x + 3 is a factor of p( x ) (2) (b) Simplify the expression (4) (Total 6 marks)

(i) Find. (3) (ii) Show that the x -coordinates of any stationary points of the curve satisfy the equation x^3 − 4 x + 15 = 0 (1) (iii) Use the results above to show that the only stationary point of the curve occurs when x = −3. (2) (iv) Find the value of when x = −3. (3) (v) Hence determine, with a reason, whether the curve has a maximum point or a minimum point when x = −3. (1) (Total 14 marks) Q8. The polynomial f( x ) is defined by f( x )= 4 x^3 − 11 x − 3. (a) Use the Factor Theorem to show that (2 x + 3) is a factor of f( x ). (2) (b) Write f( x ) in the form (2 x + 3)( ax^2 + bx + c ), where a , b and c are integers. (2) (c) (i) Show that the equation 2 cos 2 θ sin θ + 9 sin θ + 3 = 0 can be written as 4 x^3 − 11 x − 3 = 0, where x = sin θ. (3) (ii) Hence find all solutions of the equation 2 cos 2 θ sin θ + 9 sin θ + 3 = 0 in the interval 0° < θ < 360°, giving your solutions to the nearest degree. (4) (Total 11 marks) Q9. (a) Use the Factor Theorem to show that 4 x − 3 is a factor of 16 x^3 + 11 x − 15 (2) (b) Given that x = cos θ , show that the equation

27 cos θ cos 2 θ + 19 sin θ sin 2 θ − 15 = 0 can be written in the form 16 x^3 + 11 x − 15 = 0 (4) (c) Hence show that the only solutions of the equation 27 cos θ cos 2 θ + 19 sin θ sin 2 θ − 15 = 0 are given by cos θ =. (4) (Total 10 marks) Q10. The polynomial p( x ) is given by p( x ) = x^3 + 2 x^2 – 5 x – 6 (a) (i) Use the Factor Theorem to show that x + 1 is a factor of p( x ). (2) (ii) Express p( x ) as the product of three linear factors. (3) (b) Verify that p(0) > p(1). (2) (c) Sketch the curve with equation y = x^3 + 2 x^2 – 5 x – 6, indicating the values where the curve crosses the x -axis. (3) (Total 10 marks) Q11. (a) (i) Express in the form. (2) (ii) Find d x. (2) (b) (i) Given that 4 x^3 + 5 x − 2 = (2 x + 1) (2 x^2 + px + q ) + r find the values of the constants p , q and r.

Q14. The polynomial p( x ) is given by p( x ) = x^3 – 2 x^2 + 3. (a) Use the Factor Theorem to show that x + 1 is a factor of p( x ). (2) (b) (i) Express p( x ) = x^3 – 2 x^2 + 3 in the form ( x + 1)( x^2 + bx + c ), where b and c are integers. (2) (ii) Hence show that the equation p( x ) = 0 has exactly one real root. (2) (Total 6 marks) Q15. The polynomial f( x ) is defined by f( x ) = 4 x^3 – 13 x + 6. (a) Find f(–2). (1) (b) Use the Factor Theorem to show that 2 x – 3 is a factor of f( x ). (2) (c) Simplify (4) (Total 7 marks) Q16. (a) Expand (5 + h )^3. (1) (b) A curve has equation y = x^3 – x^2. (i) Find the gradient of the line passing through the point (5, 100) and the point on the curve for which x = 5 + h. Give your answer in the form p + qh + rh^2 where p, q and r are integers. (4) (ii) Show how the answer to part (b)(i) can be used to find the gradient of the curve at the point (5, 100). State the value of this gradient. (2) (Total 7 marks) Q17.

The polynomial p( x ) is given by p( x ) = x^3 – 13 x – 12. (a) Use the Factor Theorem to show that x + 3 is a factor of p( x ). (2) (b) Express p( x ) as the product of three linear factors. (3) (Total 5 marks) Q18. The polynomial f( x ) is defined by f( x ) = 15 x^3 + 19 x^2 – 4. (a) (i) Find f(–1). (1) (ii) Show that (5 x – 2) is a factor of f( x ). (2) (b) Simplify giving your answer in a fully factorised form. (5) (Total 8 marks) Q19. The expression can be written in the form , where A and B are constants. (a) Find the values of A and B. (4) (b) Hence find. (4) (Total 8 marks) Q20. The polynomial p( x ) is given by p( x ) = x^3 + 7 x^2 + 7 x – 15 (a) (i) Use the Factor Theorem to show that x + 3 is a factor of p( x ).

point Q (2, 0). (i) Find the gradient of the curve C at the point Q. (4) (ii) Hence find an equation of the tangent to the curve C at the point Q. (2) (iii) Find. (3) (iv) Hence find the area of the shaded region bounded by the curve C and the coordinate axes. (2) (Total 15 marks) Q24. The polynomial f( x ) is defined by f( x ) = 4 x^3 – 7 x – 3. (a) Find f(–1). (1) (b) Use the Factor Theorem to show that 2 x + 1 is a factor of f( x ). (2) (c) Simplify the algebraic fraction. (3) (Total 6 marks) Q25. (a) (i) Express in the form , where A and B are integers. (2) (ii) Hence find.

(2) (b) (i) Express in the form , where P , Q and R are constants. (5) (ii) Hence find. (4) (Total 13 marks) Q26. (a) The polynomial p( x ) is given by p( x ) = x^3 – x + 6. (i) Use the Factor Theorem to show that x + 2 is a factor of p( x ). (2) (ii) Express p( x ) = x^3 – x + 6 in the form ( x + 2)( x^2 + bx + c ), where b and c are integers. (2) (iii) The equation p( x ) = 0 has one root equal to –2. Show that the equation has no other real roots. (2) (b) The curve with equation y = x^3 – x + 6 is sketched below. The curve cuts the x -axis at the point A (–2, 0) and the y -axis at the point B. (i) State the y -coordinate of the point B. (1) (ii) Find. (5) (iii) Hence find the area of the shaded region bounded by the curve y = x^3 – x + 6

Q29. (a) The polynomial f( x ) is defined by f( x ) = 2 x^3 + 3 x^2 – 18 x + 8. (i) Use the Factor Theorem to show that (2 x – 1) is a factor of f( x ). (2) (ii) Write f( x ) in the form (2 x – 1)( x^2 + px + q ), where p and q are integers. (2) (iii) Simplify the algebraic fraction. (2) (b) Express the algebraic fraction in the form A + , where A , B and C are integers. (4) (Total 10 marks) Q30. (a) (i) Use the Factor Theorem to show that x + 2 is a factor of p( x ). (2) (ii) Express p( x ) as the product of linear factors. (3) (b) (i) The curve with equation y = x^3 + x^2 – 8 x – 12 passes through the point (0, k ). State the value of k. (1) (ii) Sketch the graph of y = x^3 + x^2 – 8 x – 12, indicating the values of x where the curve touches or crosses the x -axis. (3) (Total 9 marks) Q31. The polynomial f( x ) is defined by f( x ) = 27 x^3 – 9 x + 2. (a) Show that. (1) (b) Express f( x ) as a product of three linear factors. (4) (c) Simplify

(2) (Total 7 marks) Q32. The polynomial p( x ) is given by p( x ) = x^3 – 4 x^2 – 7 x + k where k is a constant. (a) (i) Given that x + 2 is a factor of p( x ), show that k = 10. (2) (ii) Express p( x ) as the product of three linear factors. (3) (b) Sketch the curve with equation y = x^3 – 4 x^2 – 7 x + 10, indicating the values where the curve crosses the x -axis and the y -axis. (You are not required to find the coordinates of the stationary points.) (4) (Total 9 marks) Q33. The polynomial f( x ) is defined by f( x ) = 2 x^3 – 7 x^2 + 13. (a) The polynomial g( x ) is defined by g( x ) = 2 x^3 – 7 x^2 + 13 + d , where d is a constant. Given that (2 x – 3) is a factor of g( x ), show that d = −4. (2) (b) Express g( x ) in the form (2 x – 3)( x^2 + ax + b ). (2) (Total 4 marks) Q34. (a) (i) Express in the form , where A and B are integers. (2) (ii) Hence find d x. (2) (b) (i) Express in the form where P and Q are integers.

Mark schemes Q1. Marking Instructions AO Marks Typical Solution Expands the bracket and obtains any correct form (PI) AO1.1b B f(3) = 36 − 36 − 3 + c = 2 c = 5 Integrates at least one of ‘their’ terms correctly AO1.1a M Substitutes x = 3 into ‘their’ integrated expression and equates to 2 AO1.1a M Obtains completely correct expression, must be explicitly stated CAO (ACF) AO1.1b A Total 4 marks Q2. Marking Instructions AO Marks Typical Solution (a) Begins a proof using a valid method Eg. Factor theorem, algebraic division, multiplication of correct factors AO1.1a M = 0 ∴ 2 x + 1 is a factor of p( x ) Constructs rigorous mathematical proof. To achieve this mark: Factor theorem the student must clearly substitute and state that p(−1/2) = 0 and clearly state that this implies that 2 x + 1 is a factor Algebraic division OR Multiplication of correct factors The method must be completely correct with a concluding statement

AO2.1 R

(b) Obtains quadratic factor PI^ AO1.1a M1 (^) p( x ) = (2 x + 1)(15 x^2 − 11 x + 2)

Obtains second linear = (2 x + 1)(5 x − 2)(3 x − 1)

factor AO1.1b A

Writes p( x ) as the product

of the correct three linear factors. NMS correct answer 3/ AO1.1b A (c) Rearranges to achieve a

cubic equation in sec x (or

cos x )

AO3.1a M

⇒ 30 sec^2 x + 2 cos x = 7 sec x +

⇒ 30 sec^3 x + 2 = 7 sec^2 x + 7 sec

x

30 sec^3 x − 7 sec^2 x − 7 sec x + 2

⇒ (2 sec x + 1)(5 sec x − 2)(

sec x − 1) = 0

These values do not fall within the

range of sec x as they are

between −1 and 1 has no real solutions. Equates to zero and uses result from (b) or factorises AO1.1a M Deduces that if solutions exist they must be of the

form sec x = −½, sec x =

1/3 or sec x = 2/5 OE

AO2.2a A Explains that the range of

sec x is (−∞,−1]∪[1,∞) OE

OE for cos x

AO2.4 E

Completes argument explaining that there cannot be any real solutions as values are outside of the function’s range

AO2.1 R

Total 10 marks

Q3.

Marking Instructions AO Marks Typical Solution Circles correct answer AO1.1b B1 9 Total 1 mark

Q4.

Marking Instructions AO Marks Typical Solution (a) Demonstrates p(−3) = 0 AO1.1b B1 p(−3) = 2(−3)^3 + 7(−3)^2 + 2(−3)− 3 = −54 + 63 − 6 − 3 = 0

p(−3) = 0 ⇒ x + 3 is a

factor Constructs rigorous mathematical proof (to achieve this mark, the student must clearly calculate and state that p(−3) = 0 and clearly state

AO2.1 R

A

[p( x )= ] ( x − 3)( x − 3)( x + 2) or [p(x)=] (x − 3)^2 (x + 2) must see product of factors A 3 (b) cubic curve with one maximum and one minimum M meeting x-axis at −2 and touching x-axis at 3 A Final A1 is dependent on previous A1 and can be withheld if curve has very poor curvature beyond x = 3, V shape at x = 3 etc graph as shown, going beyond x = −2 but condone max on or to right of y -axis A 3 [8] Q6. (a) Evaluate , not long division. M = − A 2 (b) (i) g = 0 ⇒ −3 + d = 0 Or f + d = 0 d = 3 ⇒ g( x ) = 2 x^3 + 2 x^2 − 8 x − 7 + 3

g( x ) = 2 x^3 +2 x^2 − 8 x − 4 All steps seen with conclusion AG B Allow verification with − = 0 seen, and conclusion; therefore factor 1 (ii) g( x ) = 2 x^3 + 2 x^2 − 8 x − 4 = (2 x + 1)( x^2 − 4) = (2 x + 1)( x + 2)( x − 2) a = − B 1 (iii) 2 x^3 − 3 x^2 − 2 x = x (2 x + 1)( x − 2) Clear attempt to factorise denominator; 3 factors needed. M At least one correct factor cancelled m CSO part (a)(iii) NMS is 0 / 3 A 3 Alternative: (M1) = 1 + (A1) =