Solved Examples on Polynomial Functions (All)

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For SAT Students
Any question labeled SAT-C is a question that allows a calculator.
Any question labeled SAT-NC is a question that does not allow a calculator.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.
Unless specified otherwise, any question labeled JAMB is a question from JAMB Physics

For WASSCE Students: Unless specified otherwise:
Any question labeled WASCCE is a question from WASCCE Physics
Any question labeled WASSCE-FM is a question from the WASSCE Further Mathematics/Elective Mathematics

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Show all work.
Check your solution for any question involving the factoring of quadratic trinomials. (Specifically for my Students.)
Assume all expressions to be polynomials

(1.) HSCE: Mathematics A: Paper 2: MA1 A polynomial is given by $f(x) = x^3 + 2x^2 - x - 2$

(i) Fully factorise f(x) by the grouping method and abstraction of the highest common multiple.

(ii) Hence solve the polynomial equation $f(x) = x^3 + 2x^2 - x - 2 = 0$

$ (i) \\[3ex] f(x) \\[3ex] = x^3 + 2x^2 - x - 2 \\[3ex] = (x^3 + 2x^2) - (x - 2) \\[3ex] = x^2(x + 2) - 1(x + 2) \\[3ex] = (x + 2)(x^2 - 1) \\[3ex] = (x + 2)(x^2 - 1^2) \\[3ex] = (x + 2)(x + 1)(x - 1) \\[3ex] (ii) \\[3ex] f(x) = 0 \\[3ex] x^3 + 2x^2 - x - 2 = 0 \\[3ex] (x + 2)(x + 1)(x - 1) = 0 \\[3ex] x + 2 = 0 \;\;\;OR\;\;\; x + 1 = 0 \;\;\;OR\;\;\; x - 1 = 0 \\[3ex] x = -2 \;\;\;OR\;\;\; x = -1 \;\;\;OR\;\;\; x = 1 $

(2.) WASSCE-FM The function $f(x) = px^2 + qx + r$, where p, q and r are constants.
If f(1) = 0, f(-1) = 4 and f(2) = 7, find the:

(i) values values of p, q and r

(ii) factors of f(x)

$ f(x) = px^2 + qx + r \\[3ex] f(1) = p(1)^2 + q(1) + r = 0 \\[3ex] p + q + r = 0 ...eqn.(1) \\[3ex] f(-1) = p(-1)^2 + q(-1) + r = 4 \\[3ex] p - q + r = 4 ...eqn.(2) \\[3ex] f(2) = p(2)^2 + q(2) + r = 7 \\[3ex] 4p + 2q + r = 7 ...eqn.(3) \\[3ex] eqn.(2) - eqn.(1) \implies \\[3ex] -2q = 4 \\[3ex] q = \dfrac{4}{-2} \\[5ex] q = -2 \\[3ex] eqn.(3) - eqn.(1) \implies \\[3ex] 3p + q = 7 \\[3ex] 3p = 7 - q \\[3ex] 3p = 7 - (-2) \\[3ex] 3p = 9 \\[3ex] p = \dfrac{9}{3} \\[5ex] p = 3 \\[3ex] Substitute\;\;the\;\;values\;\;of\;\;q\;\;and\;\;p\;\;in\;\;eqn.(1) \\[3ex] 3 + -2 + r = 0 \\[3ex] 1 + r = 0 \\[3ex] r = -1 \\[3ex] (ii) \\[3ex] \implies \\[3ex] f(x) = 3x^2 + (-2)x + (-1) \\[3ex] f(x) = 3x^2 - 2x - 1 \\[3ex] 3x^2(-1) = -3x^2 \\[3ex] Factors\;\;are:\;\; -3x \;\;and\;\; x \\[3ex] f(x) = 3x^2 - 3x + x - 1 \\[3ex] f(x) = 3x(x - 1) + 1(x - 1) \\[3ex] f(x) = (x - 1)(3x + 1) $

(3.) NSC Given: $f(x) = x^3 - 2x^2 - 7x - 4$
(3.1) Write down the y-intercept of f
(3.2) Show that x - 4 is a factor of f
(3.3) Determine the x-intercepts of f
(3.4) Determine the coordinates of the turning points of f
(3.5) Sketch the graph of f on the ANSWER SHEET provided. Clearly show ALL the intercepts with the axes and the turning points.
(3.6) Determine the value(s) of x for which the graph of f is decreasing.

$ f(x) = x^3 - 2x^2 - 7x - 4 \\[3ex] (3.1) \\[3ex] f(0) = 0^3 - 2(0)^2 - 7(0) - 4 \\[3ex] = 0 - 0 - 0 - 4 \\[3ex] = -4 \\[3ex] y-intercept = (0, -4) \\[3ex] (3.2) \\[3ex] Factor:\;\; x - 4 \\[3ex] Zero:\;\; x - 4 = 0 \\[3ex] x = 4 \\[3ex] f(4) = 4^3 - 2(4)^2 - 7(4) - 4 \\[3ex] = 64 - 2(16) - 28 - 4 \\[3ex] = 64 - 32 - 28 - 4 \\[3ex] = 0 \\[3ex] (3.3) \\[3ex] f(x) = 0 \\[3ex] x^3 - 2x^2 - 7x - 4 = 0 \\[5ex] 1st\;\;Factor:\;\; x - 4 \\[3ex] \begin{array}{c|c} & x^2 + 2x + 1 \\ \hline x - 4 & x^3 - 2x^2 - 7x - 4 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &x^3 - 4x^2~~~~~~~~~~~~~~~~ \\ \hline &~~2x^2 - 7x \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~2x^2 - 8x \\ \hline &~~~~~~~~~~~~~~~~~~x - 4 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~~~~~~~~~~~~~~~~~x - 4 \\ \hline &~~~~~~~~~~~~~~~~~~0 \\[5ex] \end{array} \\[5ex] Quotient = x^2 + 2x + 1 \\[3ex] x^2 + 2x + 1 = 0 \\[3ex] (x + 1)(x + 1) = 0 \\[3ex] x + 1 = 0 \\[3ex] x = -1\;(repeated) \\[3ex] x-intercepts = (4, 0),\;\;(-1, 0),\;\;and\;\;(-1, 0) \\[3ex] (3.4) \\[3ex] f'(x) = 3x^2 - 4x - 7 \\[3ex] f'(x) = 0 \\[3ex] \implies \\[3ex] 3x^2 - 4x - 7 = 0 \\[3ex] 3x^2 + 3x - 7x - 7 = 0 \\[3ex] 3x^2 + 3x - 7x - 7 = 0 \\[3ex] 3x(x + 1) - 7(x + 1) = 0 \\[3ex] (x + 1)(3x - 7) = 0 \\[3ex] x + 1 = 0 \;\;\;OR\;\;\; 3x - 7 = 0 \\[3ex] x = -1 \;\;\;OR\;\;\; 3x = 7 \\[3ex] x = -1 \;\;\;OR\;\;\; x = \dfrac{7}{3} \\[5ex] Critical\;\;Values: x = -1 \;\;\;OR\;\;\; x = \dfrac{7}{3} \\[5ex] f(-1) = (-1)^3 - 2(-1)^2 - 7(-1) - 4 \\[3ex] = -1 - 2(1) + 7 - 4 \\[3ex] = -1 - 2 + 7 - 4 \\[3ex] = 0 \\[3ex] f\left(\dfrac{7}{3}\right) = \left(\dfrac{7}{3}\right)^3 - 2\left(\dfrac{7}{3}\right)^2 - 7\left(\dfrac{7}{3}\right) - 4 \\[5ex] = \dfrac{343}{27} - 2\left(\dfrac{49}{9}\right) - \dfrac{49}{3} - 4 \\[5ex] = \dfrac{343}{27} - \dfrac{98}{9} - \dfrac{49}{3} - 4 \\[5ex] = \dfrac{343 - 294 - 441 - 108}{27} \\[5ex] = -\dfrac{500}{27} \\[5ex] Turning\;\;Points:\;\; (-1,0) \;\;\;AND\;\;\; \left(\dfrac{7}{3},-\dfrac{500}{27}\right) \\[5ex] Turning\;\;Points:\;\; (-1,0) \;\;\;AND\;\;\; (2.333, -18.519) \\[5ex] $ (3.5)
The graph of f is:

$ (3.6) \\[3ex] x \downarrow \;\;for\;\; \left(-1, \dfrac{7}{3}\right) $

(4.) JAMB If x + 2 and x − 1 are factors of the expression $lx + 2kx^2 + 24$, find the values of l and k

$ A.\;\; l = -6,\;\;\;k = -9 \\[3ex] B.\;\; l = -2,\;\;\;k = 1 \\[3ex] C.\;\; l = -2,\;\;\;k = -1 \\[3ex] D.\;\; l = 0,\;\;\;k = 1 \\[3ex] E.\;\; l = 6,\;\;\;k = 0 \\[3ex] $

$ f(x) = lx^3 + 2kx^2 + 24 \\[3ex] 1st\;\;Factor:\;\; x + 2 \\[3ex] 1st:\;\;Zero:\;\; x + 2 = 0 \\[3ex] x = -2 \\[3ex] f(-2) = l(-2)^3 + 2k(-2)^2 + 24 \\[3ex] = -8l + 8k + 24 \\[3ex] f(-2) = 0 ...Factor\;\;Theorem \\[3ex] \implies \\[3ex] -8l + 8k + 24 = 0 \\[3ex] -l + k + 3 = 0 \\[3ex] -l + k = -3...eqn.(1) \\[3ex] 2nd\;\;Factor:\;\; x - 1 \\[3ex] 2nd:\;\;Zero:\;\; x - 1 = 0 \\[3ex] x = 1 \\[3ex] f(1) = l(1)^3 + 2k(1)^2 + 24 \\[3ex] = l + 2k + 24 \\[3ex] f(1) = 0 ...Factor\;\;Theorem \\[3ex] \implies \\[3ex] l + 2k + 24 = 0 \\[3ex] l + 2k = -24 \\[3ex] l + 2k = -24...eqn.(2) \\[3ex] eqn.(1) + eqn.(2) \implies \\[3ex] 3k = -27 \\[3ex] k = -\dfrac{27}{3} \\[5ex] k = -9 \\[3ex] Substitute\;\;k = -9\;\;into\;\;eqn.(2) \\[3ex] l + 2(-9) = -24 \\[3ex] l - 18 = -24 \\[3ex] l = -24 + 18 \\[3ex] l = -6 \\[3ex] l = -6,\;\;\;k = -9 $

(6.) ICSE Using the Factor Theorem, show that $(x - 2)$ is a factor of $x^3 + x^2 - 4x - 4$.
Hence factorise the polynomial completely.

$ Factor:\;\; x - 2 \\[3ex] Zero: x - 2 = 0 \\[3ex] x = 2 \\[3ex] f(x) = x^3 + x^2 - 4x - 4 \\[3ex] f(2) = 2^3 + 2^2 - 4(2) - 4 \\[3ex] = 8 + 4 - 8 - 4 \\[3ex] = 0 \\[3ex] \therefore x - 2 \;\;is\;\;a\;\;factor\;\;of\;\;x^3 + x^2 - 4x - 4 \\[3ex] 1st\;\;Factor:\;\; x - 2 \\[3ex] \begin{array}{c|c} & x^2 + 3x + 2~~~~~~~~~~ \\ \hline x - 2 & x^3 + x^2 - 4x - 4 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &x^3 - 2x^2~~~~~~~~~~~~~ \\ \hline &~~3x^2 - 4x \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~3x^2 - 6x \\ \hline &~~~~~~~~~~~~~~~~~~2x - 4 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~~~~~~~~~~~~~~~~~2x - 4 \\ \hline &~~~~~~~~~~~~~~~~~~~~0 \\[5ex] \end{array} \\[5ex] Quotient = x^2 + 3x + 2 \\[3ex] Quotient = (x + 2)(x + 1) \\[3ex] \therefore factors\;\;of\;\;x^3 + x^2 - 4x - 4 = (x - 2)(x + 2)(x + 1) $

(8.) WASSCE-FM The polynomial $x^3 + qx^2 + rx + 9$, where q and r are constants, has (x + 1) as a factor and has a remainder −17 when divided by (x + 2)
Find the values of q and r

$ \underline{Factor\;\;Theorem} \\[3ex] Factor = x + 1 \\[3ex] Zero:\;\; x + 1 = 0 \\[3ex] x = -1 \\[3ex] f(-1) = (-1)^3 + q(-1)^2 + r(-1) + 9 = 0 \\[3ex] -1 + q - r + 9 = 0 \\[3ex] q - r + 8 = 0 \\[3ex] q - r = -8 ...eqn.(1) \\[3ex] \underline{Remainder\;\;Theorem} \\[3ex] Remainder = -17 \\[3ex] f(-2) = (-2)^3 + q(-2)^2 + r(-2) + 9 = -17 \\[3ex] -8 + 4q - 2r + 9 = -17 \\[3ex] 4q - 2r + 1 = -17 \\[3ex] 4q - 2r = -17 - 1 \\[3ex] 4q - 2r = -18 \\[3ex] 2q - r = -9 ...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \implies \\[3ex] q = -1 \\[3ex] Substitute\;\;q = -1 \;\;into\;\;eqn.(1) \\[3ex] -1 - r = -8 \\[3ex] -1 + 8 = r \\[3ex] r = 7 $

(10.) ICSE Use Remainder theorem to factorize the following polynomial:

$ 2x^3 + 3x^2 - 9x - 10 \\[3ex] $

$ f(x) = 2x^3 + 3x^2 - 9x - 10 \\[3ex] Test:\;\; x = 2 \\[3ex] f(2) = 2(2)^3 + 3(2)^2 - 9(2) - 10 \\[3ex] = 16 + 12 - 18 - 10 \\[3ex] = 0 \\[3ex] \implies \\[3ex] Zero:\;\; x = 2 \\[3ex] 1st\;\;Factor:\;\; x - 2 \\[3ex] \begin{array}{c|c} & 2x^2 + 7x + 5 \\ \hline x - 2 & 2x^3 + 3x^2 - 9x - 10 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &2x^3 - 4x^2~~~~~~~~~~~~~~~~ \\ \hline &~~7x^2 - 9x \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~~~7x^2 - 14x \\ \hline &~~~~~~~~~~~~~~~~~~~~~5x - 10 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~~~~~~~~~~~~~~~~~~~~5x - 10 \\ \hline &~~~~~~~~~~~~~~~~~~~~~0 \\[5ex] \end{array} \\[5ex] Quotient = 2x^2 + 7x + 5 \\[3ex] \underline{Factor\;\;Quadratic\;\;Trinomial} \\[3ex] (2x^2)(5) = 10x^2 \\[3ex] Factors\;\;are:\;\;2x\;\;and\;\;5x \\[3ex] \implies \\[3ex] 2x^2 + 2x + 5x + 5 \\[3ex] 2x(x + 1) + 5(x + 1) \\[3ex] (x + 1)(2x + 5) \\[3ex] \therefore 2x^3 + 3x^2 - 9x - 10 = (x - 2)(x + 1)(2x + 5) $

(12.) MEHA Calculate the value of k given that x + 2 is a factor of $f(x) = x^3 + 8x^2 + 7x + k$

$ Factor:\;\; x + 2 \\[3ex] Zero:\;\; x + 2 = 0 \\[3ex] x = -2 \\[3ex] f(-2) = 0 .. eqn.(1) ...Factor\;\;Theorem \\[3ex] f(x) = x^3 + 8x^2 + 7x + k \\[3ex] f(-2) = (-2)^3 + 8(-2)^2 + 7(-2) + k \\[3ex] = -8 + 32 - 14 + k \\[3ex] = 10 + k...eqn.(2) \\[3ex] f(-2) = f(-2) \impies eqn.(2) = eqn.(1) \\[3ex] 10 + k = 0 \\[3ex] k = -10 $

(14.) ICSE What must be subtracted from $16x^3 - 8^2 + 4x + 7$ so that the resulting expression has $2x + 1$ as a factor?

$ Factor:\;\;2x + 1 \\[3ex] Zero:\;\; 2x + 1 = 0 \\[3ex] 2x = -1 \\[3ex] x = -\dfrac{1}{2} \\[5ex] Let:\;\;f(x) = 16x^3 - 8^2 + 4x + 7 \\[3ex] f\left(-\dfrac{1}{2}\right) = 16\left(-\dfrac{1}{2}\right)^3 - 8\left(-\dfrac{1}{2}\right)^2 + 4\left(-\dfrac{1}{2}\right) + 7 \\[5ex] = 16\left(-\dfrac{1}{8}\right) - 8\left(\dfrac{1}{4}\right) - 2 + 7 \\[5ex] = -2 - 2 - 2 + 7 \\[3ex] = 1 \\[3ex] $ When $-\dfrac{1}{2}$ is substituted in the polynomial, the remainder is 1
If we subtract that remainder (which is 1) from the polynomial, then $-\dfrac{1}{2}$ would be a zero
So, that $2x + 1$ would be a factor
Therefore, 1 must be subtracted from $16x^3 - 8^2 + 4x + 7$ so that the resulting expression: $16x^3 - 8^2 + 4x + 6$ has $2x + 1$ as a factor.

(16.) JAMB If (x − 2) and (x + 1) are factors of the expression $x^3 + px^2 + qx + 1$, what is the sum of p and q?

$ A.\;\; 9 \\[3ex] B.\;\; -3 \\[3ex] C.\;\; 3 \\[3ex] D.\;\; \dfrac{17}{3} \\[5ex] E.\;\; \dfrac{2}{3} \\[5ex] $

$ f(x) = x^3 + px^2 + qx + 1 \\[3ex] 1st\;\;Factor:\;\; x - 2 \\[3ex] 1st:\;\;Zero:\;\; x - 2 = 0 \\[3ex] x = 2 \\[3ex] f(2) = 2^3 + p(2)^2 + q(2) + 1 \\[3ex] = 8 + 4p + 2q + 1 \\[3ex] = 4p + 2q + 9 \\[3ex] f(2) = 0 ...Factor\;\;Theorem \\[3ex] \implies \\[3ex] 4p + 2q + 9 = 0 //[3ex] 4p + 2q = -9 ...eqn(1) \\[3ex] 2nd\;\;Factor:\;\; x + 1 \\[3 ex] 2nd:\;\;Zero:\;\; x + 1 = 0 \\[3ex] x = -1 \\[3ex] f(-1) = (-1)^3 + p(-1)^2 + q(-1) + 1 \\[3ex] = -1 + p - q + 1 \\[3ex] = p - q \\[3ex] f(1) = 0 ...Factor\;\;Theorem \\[3ex] \implies \\[3ex] p - q = 0 ...eqn.(2) \\[3ex] eqn.(1) - 4 * eqn.(2) \implies \\[3ex] (4p + 2q) - 4(p - q) = -9 - 4(0) \\[3ex] 4p + 2q - 4p + 4q = -9 - 0 \\[3ex] 6q = -9 \\[3ex] q = -\dfrac{9}{6} \\[5ex] q = -\dfrac{3}{2} \\[5ex] Substitute\;\;q = -\dfrac{3}{2}\;\;into\;\;eqn.(2) \\[5ex] p - -\dfrac{3}{2} = 0 \\[5ex] p + \dfrac{3}{2} = 0 \\[5ex] p = -\dfrac{3}{2} \\[5ex] p + q \\[3ex] = -\dfrac{3}{2} + -\dfrac{3}{2} \\[5ex] = -\dfrac{3}{2} - \dfrac{3}{2} \\[5ex] = -\dfrac{6}{2} \\[5ex] = -3 $

(18.) GCSE (a) s and t are positive integers.
(x + s)(x − t) is expanded and simplified.
The answer is x² + kx − 40 where k is a positive integer.
Work out the smallest possible value of k

(b) Faisal tries to solve (x + 2)(x − 7) = 0
Here is his working.

Give a reason why his answer is wrong.

$ (a) \\[3ex] (x + s)(x - t) \\[3ex] x^2 - xt + sx - st \\[3ex] x^2 + x(-t + s) - st \\[3ex] Compare\;\;to:\;\; x^2 + kx - 40 \\[3ex] -t + s = k \\[3ex] k = s - t \\[3ex] st = 40 \\[3ex] s \gt t \;\;because\;\;k\;\;is\;\;positive \\[3ex] Factors\;\;of\;\;40\;\;that\;\;satisfies\;\;it\;\;are: \\[3ex] 20\;\;and\;\;2 \\[3ex] 10\;\;and\;\;4 \\[3ex] 8\;\;and\;\;5 \\[3ex] 20 - 2 = 18 \\[3ex] 10 - 4 = 6 \\[3ex] 8 - 5 = 3 \\[3ex] 3 \lt 6 \lt 18 \\[3ex] \therefore the\;\;smallest\;\;possible\;\;value\;\;of\;\;k = 3 \\[3ex] $ (b)
A reason why his answer is wrong is:
First Binomial: If x + 2 = 0, then x = −2 rather than 2

(20.) ICSE Using the Remainder Theorem, find the remainders obtained when $x^3 + (kx + 8)x + k$ is divided by $x + 1$ and $x - 2$
Hence find k if the sum of the two remainders is 1.

$ f(x) = x^3 + (kx + 8)x + k \\[3ex] f(x) = x^3 + x(kx + 8) + k \\[3ex] \underline{Remainder\;\;Theorem} \\[3ex] 1st\;\;Non\;\;Factor:\;\; x + 1 \\[3ex] Non-Zero:\;\; x + 1 = 0 \\[3ex] x = -1 \\[3ex] 1st\;\;Remainder = f(-1) = (-1)^3 + -1[k(-1) + 8] + k \\[3ex] = -1 + -1(-k + 8) + k \\[3ex] = -1 + k - 8 + k \\[3ex] = 2k - 9 ...eqn.(1) \\[3ex] 1st\;\;Non\;\;Factor:\;\; x - 2 \\[3ex] Non-Zero:\;\; x - 2 = 0 \\[3ex] x = 2 \\[3ex] 2nd\;\;Remainder = f(2) = 2^3 + 2[k(2) + 8] + k \\[3ex] = 8 + 2(2k + 8) + k \\[3ex] = 8 + 4k + 16 + k \\[3ex] = 5k + 24 ...eqn.(2) \\[3ex] Sum\;\;of\;\;Remainders = 1 \\[3ex] \implies \\[3ex] eqn.(1) + eqn.(2) = 1 \\[3ex] 2k - 9 + 5k + 24 = 1 \\[3ex] 7k + 15 = 1 \\[3ex] 7k = 1 - 15 \\[3ex] 7k = -14 \\[3ex] k = -\dfrac{14}{7} \\[5ex] k = -2 $

Top

(22.) USSCE: Advance Mathematics Paper 2: MA2 Find constants a and b, given that the polynomial $P(x) = x^3 + ax^2 -bx - 10$ is divisible by $x + 1$ and $x - 2$

$ P(x) = x^3 + ax^2 -bx - 10 \\[3ex] 1st\;\;Factor:\;\; x + 1 \\[3ex] 1st\;\;Zero:\;\; x + 1 = 0 \\[3ex] x = -1 \\[3ex] P(-1) = (-1)^3 + a(-1)^2 -b(-1) - 10 \\[3ex] 0 = -1 + a + b - 10 \\[3ex] a + b - 11 = 0 \\[3ex] a + b = 11 ...eqn.(1) \\[3ex] 2nd\;\;Factor:\;\; x - 2 \\[3ex] 2nd\;\;Zero:\;\; x - 2 = 0 \\[3ex] x = 2 \\[3ex] P(2) = 2^3 + a(2)^2 -b(2) - 10 \\[3ex] 0 = 8 + 4a - 2b - 10 \\[3ex] 4a - 2b - 2 = 0 \\[3ex] 4a - 2b = 2 ...eqn.(2) \\[3ex] 2 * eqn.(1) \\[3ex] 2a + 2b = 2(11) \\[3ex] 2a + 2b = 22 ...eqn.(3) \\[3ex] eqn.(2) + eqn.(3) \\[3ex] 4a + 2a - 2b + 2b = 2 + 22 \\[3ex] 6a = 24 \\[3ex] a = \dfrac{24}{6} \\[5ex] a = 4 \\[3ex] Substitute\;\;a = 4 \;\;into\;\;eqn.(1) \\[3ex] 4 + b = 11 \\[3ex] b = 11 - 4 \\[3ex] b = 7 $

(24.) ICSE Using Remainder Theorem, find the value of k if on dividing $2x^3 + 3x^2 - kx + 5$ by $x - 2$, leaves a remainder 7.

$ f(x) = 2x^3 + 3x^2 - kx + 5 \\[3ex] Non-factor:\;\; x - 2 \\[3ex] Non-zero:\;\; x - 2 = 0 \\[3ex] x = 2 \\[3ex] f(2) = 7 ...Remainder\;\;Theorem ...eqn.(1) \\[3ex] f(2) = 2(2)^3 + 3(2)^2 - k(2) + 5 \\[3ex] f(2) = 16 + 12 - 2k + 5 \\[3ex] f(2) = 33 - 2k ...eqn.(2) \\[3ex] f(2) = f(2) \implies eqn.(1) = eqn.(2) \\[3ex] 7 = 33 - 2k \\[3ex] 2k = 33 - 7 \\[3ex] 2k = 26 \\[3ex] k = \dfrac{26}{2} \\[5ex] k = 13 $

(28.) ICSE Use Factor Theorem to factorise $6x^3 + 17x^2 + 4x - 12$ completely.

$ f(x) = 6x^3 + 17x^2 + 4x - 12 \\[3ex] Test\;\; x = -2 \\[3ex] f(-2) = 6(-2)^3 + 17(-2)^2 + 4(-2) - 12 \\[3ex] = -48 + 68 - 8 - 12 \\[3ex] = 0 \\[3ex] \implies \\[3ex] 1st\;\;Zero:\;\; x = -2 \\[3ex] 1st\;\;Factor = x + 2 \\[3ex] \begin{array}{c|c} & 6x^2 + 5x - 6 \\ \hline x + 2 & ~~6x^3 + 17x^2 + 4x - 12 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &6x^3 + 12x^2~~~~~~~~~~~~~~~~ \\ \hline &~~5x^2 + 4x \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~~~5x^2 + 10x \\ \hline &~~~~~~~~~~~~~~~~~~~~~-6x - 12 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~~~~~~~~~~~~~~~~~~~~-6x - 12 \\ \hline &~~~~~~~~~~~~~~~~~~~~~~~~~~~0 \\[5ex] \end{array} \\[5ex] Quotient = 6x^2 + 5x - 6 \\[3ex] \underline{Factor\;\;Quadratic\;\;Trinomial} \\[3ex] (6x^2)(-6) = -36x^2 \\[3ex] Factors\;\;are:\;\;9x\;\;and\;\;-4x \\[3ex] \implies \\[3ex] 6x^2 + 9x - 4x - 6 \\[3ex] 3x(2x + 3) - 2(2x + 3) \\[3ex] (2x + 3)(3x - 2) \\[3ex] \therefore 2x^3 + 3x^2 - 9x - 10 = (x + 2)(2x + 3)(3x - 2) $

(32.) ICSE What must be added to the polynomial $2x^3 - 3x^2 - 8x$, so that it leaves a remainder 10 when divided by $2x + 1$?

$ Factor:\;\;2x + 1 \\[3ex] Zero:\;\; 2x + 1 = 0 \\[3ex] 2x = -1 \\[3ex] x = -\dfrac{1}{2} \\[5ex] Let:\;\;f(x) = 2x^3 - 3^2 - 8x \\[3ex] f\left(-\dfrac{1}{2}\right) = 2\left(-\dfrac{1}{2}\right)^3 - 3\left(-\dfrac{1}{2}\right)^2 - 8\left(-\dfrac{1}{2}\right) \\[5ex] = 2\left(-\dfrac{1}{8}\right) - 3\left(\dfrac{1}{4}\right) + 4 \\[5ex] = -\dfrac{1}{4} - \dfrac{3}{4} + \dfrac{16}{4} \\[5ex] = -\dfrac{12}{4} \\[5ex] = -3 \\[3ex] $ When $-\dfrac{1}{2}$ is substituted in the polynomial, the remainder is -3
But we want a remainder of 10
So, what must be added to -3 to give 10?
Let that thing be p

$ -3 + p = 10 \\[3ex] p = 10 + 3 \\[3ex] p = 13 \\[3ex] $ Therefore, 13 must be added to $2x^3 - 3x^2 - 8x$ so that when the new polynomial: $2x^3 - 3x^2 - 8x + 13$ is divided by $2x +1$, the remainder is 10

(36.) ICSE Show that $(x - 1)$ is a factor of $x^3 - 7x^2 + 14x - 8$
Hence, completely factorise the above expression.

$ f(x) = x^3 - 7x^2 + 14x - 8 \\[3ex] Factor:\;\; x - 1 \\[3ex] Zero:\;\; x - 1 = 0 \\[3ex] x = 1 \\[3ex] f(1) = 1^3 - 7(1)^2 + 14(1) - 8 \\[3ex] = 1 - 7 + 14 - 8 \\[3ex] = 0 \\[3ex] \therefore x - 1 \;\;is\;\;a\;\;factor\;\;of\;\;x^3 - 7x^2 + 14x - 8 \\[3ex] 1st\;\;Factor:\;\; x - 1 \\[3ex] \begin{array}{c|c} & x^2 - 6x + 8~~~~~~~~~~~~~~ \\ \hline x - 1 & x^3 - 7x^2 + 14x - 8 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &x^3 - x^2~~~~~~~~~~~~~~~~~~~ \\ \hline &-6x^2 + 14x \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &-6x^2 + 6x \\ \hline &~~~~~~~~~~~~~~~~~~~~~8x - 8 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~~~~~~~~~~~~~~~~~~~~8x - 8 \\ \hline &~~~~~~~~~~~~~~~~~~~~~~~~0 \\[5ex] \end{array} \\[5ex] Quotient = x^2 - 6x + 8 \\[3ex] Quotient = (x - 2)(x - 4) \\[3ex] \therefore factors\;\;of\;\;x^3 - 7x^2 + 14x - 8 = (x - 1)(x - 2)(x - 4) $

(39.) JAMB The solutions of $x^2 - 2x - 1 = 0$ are the points of intersection of two graphs.
If one of the graphs is $y = 2 + x - x^2$, find the second graph.

$ A.\;\; y = 1 - x \\[3ex] B.\;\; y = 1 + x \\[3ex] C.\;\; y = x - 1 \\[3ex] D.\;\; y = 3x + 3 \\[3ex] $

$ solution\;\;of\;\;polynomial\;\;equation = second\;\;polynomial - first\;\;polynomial \\[3ex] x^2 - 2x - 1 = second\;\;polynomial - (2 + x - x^2) \\[3ex] x^2 - 2x - 1 + (2 + x - x^2) = second\;\;polynomial \\[3ex] second\;\;polynomial = x^2 - 2x - 1 + 2 + x - x^2 \\[3ex] second\;\;polynomial = -x + 1 \\[3ex] y = -x + 1 \\[3ex] y = 1 - x \\[3ex] $ Let us confirm with a graph.

Top

(42.) ICSE If $(x - 2)$ is a factor of $2x^3 - x^2 - px - 2$
(i) find the value of p
(ii) with the value of p, factorize the above expression completely.

$ 1st\;\;Factor:\;\; x - 2 \\[3ex] Zero:\;\; x - 2 = 0 \\[3ex] x = 2 \\[3ex] f(x) = 2x^3 - x^2 - px - 2 \\[3ex] f(2) = 2(2)^3 - 2^2 - p(2) - 2 \\[3ex] = 16 - 4 - 2p - 2 \\[3ex] = 10 - 2p \\[3ex] f(2) = 0 ...Factor\;\;Theorem \\[3ex] \implies \\[3ex] 10 - 2p = 0 \\[3ex] 10 = 2p \\[3ex] 2p = 10 \\[3ex] p = \dfrac{10}{2} \\[5ex] p = 5 \\[3ex] \therefore f(x) = 2x^3 - x^2 - 5x - 2 \\[3ex] \begin{array}{c|c} & 2x^2 + 3x + 1~~~~~~~~~ \\ \hline x - 2 & 2x^3 - x^2 - 5x - 2 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &2x^3 - 4x^2~~~~~~~~~~~~~ \\ \hline &~~~~~~3x^2 - 5x \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~~~~~3x^2 - 6x \\ \hline &~~~~~~~~~~~~~~~~~~~~~~x - 2 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~~~~~~~~~~~~~~~~~~~~~x - 2 \\ \hline &~~~~~~~~~~~~~~~~~~~~~~0 \\[5ex] \end{array} \\[5ex] Quotient = 2x^2 + 3x + 1 \\[3ex] = 2x^2 + 2x + x + 1 \\[3ex] = 2x(x + 1) + 1(x + 1) \\[3ex] = (x + 1)(2x + 1) \\[3ex] \therefore factors\;\;of\;\;2x^3 - x^2 - 5x - 2 = (x - 2)(x + 1)(2x + 1) $