CBSE Class 9 Mathematics Polynomials Worksheet Set 07

Question 1. Find the zeroes of the following polynomials:
i) x + 1
ii) 2x - 1
iii) x - √3
iv) 3x - π
v) 3x + 2
vi) √2x - 1
vii) (x - 1)(x + 2)
viii) ax + b
Answer: To find the zero of any polynomial \( p(x) \), equate \( p(x) \) to 0 and solve for \( x \):
i) \( x + 1 = 0 \)
\( \implies x = -1 \)
ii) \( 2x - 1 = 0 \)
\( \implies 2x = 1 \)
\( \implies x = \frac{1}{2} \)
iii) \( x - \sqrt{3} = 0 \)
\( \implies x = \sqrt{3} \)
iv) \( 3x - \pi = 0 \)
\( \implies 3x = \pi \)
\( \implies x = \frac{\pi}{3} \)
v) \( 3x + 2 = 0 \)
\( \implies 3x = -2 \)
\( \implies x = -\frac{2}{3} \)
vi) \( \sqrt{2}x - 1 = 0 \)
\( \implies \sqrt{2}x = 1 \)
\( \implies x = \frac{1}{\sqrt{2}} \) (or \( \frac{\sqrt{2}}{2} \))
vii) \( (x - 1)(x + 2) = 0 \)
\( \implies x - 1 = 0 \) or \( x + 2 = 0 \)
\( \implies x = 1 \) or \( x = -2 \)
viii) \( ax + b = 0 \) (where \( a \neq 0 \))
\( \implies ax = -b \)
\( \implies x = -\frac{b}{a} \)
In simple words: Set each polynomial equal to zero and isolate x using basic algebra.

Exam Tip: For linear polynomials of the form \( ax + b \), state the condition \( a \neq 0 \) when writing the zero as \( -\frac{b}{a} \).

Question 2. Verify whether ½ and -1 are zeroes of the polynomial f(x) = 2x² + x - 1
Answer: A value \( c \) is a zero of \( f(x) \) if evaluating the polynomial at \( c \) gives \( f(c) = 0 \).
Substitute \( x = \frac{1}{2} \):
\( f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right) - 1 \)
\( = 2\left(\frac{1}{4}\right) + \frac{1}{2} - 1 \)
\( = \frac{1}{2} + \frac{1}{2} - 1 = 1 - 1 = 0 \)
Since \( f\left(\frac{1}{2}\right) = 0 \), \( \frac{1}{2} \) is a zero of \( f(x) \).
Now substitute \( x = -1 \):
\( f(-1) = 2(-1)^2 + (-1) - 1 \)
\( = 2(1) - 1 - 1 = 2 - 2 = 0 \)
Since \( f(-1) = 0 \), \( -1 \) is also a zero of \( f(x) \).
Hence, both \( \frac{1}{2} \) and \( -1 \) are verified to be zeroes of \( f(x) \).
In simple words: Plug both values into the polynomial; since both give a final answer of zero, both are verified as roots.

Exam Tip: Always show the substitution step clearly before simplifying, and write a concluding sentence confirming the verification.

Question 3. If -2 is a zero of the polynomial 3x² + 2x + k, find the value of k.
Answer: Let \( p(x) = 3x^2 + 2x + k \).
Because \( -2 \) is given as a zero of \( p(x) \), we have \( p(-2) = 0 \).
Substitute \( x = -2 \) into the polynomial:
\( 3(-2)^2 + 2(-2) + k = 0 \)
\( 3(4) - 4 + k = 0 \)
\( 12 - 4 + k = 0 \)
\( 8 + k = 0 \)
\( \implies k = -8 \).
Thus, the value of \( k \) is \( -8 \).
In simple words: Since -2 makes the expression zero, substitute -2 for x and solve the linear equation for k.

Exam Tip: Watch the signs carefully: \( (-2)^2 = +4 \), whereas \( 2(-2) = -4 \).

Question 4. Find the remainder when p(x) = 4x³ - 12x² + 14x - 3 is divided by x - ½
Answer: According to the Remainder Theorem, when a polynomial \( p(x) \) is divided by \( x - c \), the remainder equals \( p(c) \).
Setting the divisor to zero:
\( x - \frac{1}{2} = 0 \implies x = \frac{1}{2} \)
Now evaluate \( p\left(\frac{1}{2}\right) \):
\( p\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 12\left(\frac{1}{2}\right)^2 + 14\left(\frac{1}{2}\right) - 3 \)
\( = 4\left(\frac{1}{8}\right) - 12\left(\frac{1}{4}\right) + 7 - 3 \)
\( = \frac{1}{2} - 3 + 7 - 3 \)
\( = \frac{1}{2} + 1 = \frac{3}{2} \)
Therefore, the required remainder is \( \frac{3}{2} \).
In simple words: Put 1/2 in place of x in the given polynomial to compute the remainder directly without long division.

Exam Tip: State the Remainder Theorem explicitly to secure method marks before writing down the numeric evaluation.

Question 5. Verify whether f(x) = 4x³ + 4x² - x - 1 is exactly divisible by 2x + 1
Answer: By the Factor Theorem, \( f(x) \) is divisible by \( 2x + 1 \) if and only if \( f\left(-\frac{1}{2}\right) = 0 \).
Find the zero of the divisor:
\( 2x + 1 = 0 \implies x = -\frac{1}{2} \)
Substitute \( x = -\frac{1}{2} \) into \( f(x) \):
\( f\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{2}\right)^3 + 4\left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right) - 1 \)
\( = 4\left(-\frac{1}{8}\right) + 4\left(\frac{1}{4}\right) + \frac{1}{2} - 1 \)
\( = -\frac{1}{2} + 1 + \frac{1}{2} - 1 = 0 \)
Because the remainder is 0, \( f(x) \) is confirmed to be exactly divisible by \( 2x + 1 \).
In simple words: Since evaluating the polynomial at x = -1/2 yields 0, the divisor divides it completely without remainder.

Exam Tip: An odd power keeps the negative sign: \( \left(-\frac{1}{2}\right)^3 = -\frac{1}{8} \).

Question 6. Find the value of 'a' so that p(x) = x³ - ax² - 13x + 15 is exactly divisible by x - 1
Answer: Applying the Factor Theorem, \( p(x) \) is divisible by \( x - 1 \) if \( p(1) = 0 \).
Substitute \( x = 1 \):
\( p(1) = (1)^3 - a(1)^2 - 13(1) + 15 = 0 \)
\( 1 - a - 13 + 15 = 0 \)
\( 3 - a = 0 \)
\( \implies a = 3 \).
Hence, the value of \( a \) is 3.
In simple words: Set x to 1, equate the whole polynomial to zero, and solve the equation for a.

Exam Tip: Combine the constant terms first: \( 1 - 13 + 15 = 3 \), then isolate \( a \).

Question 7. If x - 1 and x + 3 are factors of x³ - ax² - 13x + b, find the values of a and b.
Answer: Let \( p(x) = x^3 - ax^2 - 13x + b \).
Since \( x - 1 \) is a factor, \( p(1) = 0 \):
\( (1)^3 - a(1)^2 - 13(1) + b = 0 \)
\( 1 - a - 13 + b = 0 \)
\( -a + b - 12 = 0 \implies b - a = 12 \) ...(1)
Since \( x + 3 \) is a factor, \( p(-3) = 0 \):
\( (-3)^3 - a(-3)^2 - 13(-3) + b = 0 \)
\( -27 - 9a + 39 + b = 0 \)
\( 12 - 9a + b = 0 \implies b - 9a = -12 \) ...(2)
Subtract equation (2) from equation (1):
\( (b - a) - (b - 9a) = 12 - (-12) \)
\( 8a = 24 \)
\( \implies a = 3 \)
Substitute \( a = 3 \) into equation (1):
\( b - 3 = 12 \)
\( \implies b = 15 \).
Thus, \( a = 3 \) and \( b = 15 \).
In simple words: Form two linear equations by setting p(1) = 0 and p(-3) = 0, then solve them together to find a and b.

Exam Tip: Be careful with signs when calculating powers of negative numbers: \( (-3)^3 = -27 \) and \( (-3)^2 = +9 \).

Question 8. Find the values of a and b so that x - 1 and x + 2 are factors of f(x) = 2x³ + ax² + bx - 14
Answer: Using the Factor Theorem:
Since \( x - 1 \) is a factor of \( f(x) \), \( f(1) = 0 \):
\( 2(1)^3 + a(1)^2 + b(1) - 14 = 0 \)
\( 2 + a + b - 14 = 0 \)
\( a + b = 12 \) ...(1)
Since \( x + 2 \) is a factor of \( f(x) \), \( f(-2) = 0 \):
\( 2(-2)^3 + a(-2)^2 + b(-2) - 14 = 0 \)
\( 2(-8) + 4a - 2b - 14 = 0 \)
\( -16 + 4a - 2b - 14 = 0 \)
\( 4a - 2b - 30 = 0 \)
Divide throughout by 2:
\( 2a - b = 15 \) ...(2)
Add equation (1) and equation (2):
\( (a + b) + (2a - b) = 12 + 15 \)
\( 3a = 27 \)
\( \implies a = 9 \)
Substitute \( a = 9 \) into equation (1):
\( 9 + b = 12 \)
\( \implies b = 3 \).
Therefore, \( a = 9 \) and \( b = 3 \).
In simple words: Plug in 1 and -2 into the polynomial to get two simultaneous equations, then add them to solve for a and b.

Exam Tip: Simplifying equation (2) by dividing by 2 makes solving the system much quicker and less prone to calculation errors.

Question 9. If p(x) = ax³ + 4x² + 3x - 4 and q(x) = x³ - 4x + a leave the same remainder when divided by x - 3, find the value of a.
Answer: By the Remainder Theorem, the remainders when dividing by \( x - 3 \) are \( p(3) \) and \( q(3) \) respectively.
Calculate remainder \( R_1 = p(3) \):
\( p(3) = a(3)^3 + 4(3)^2 + 3(3) - 4 \)
\( = 27a + 4(9) + 9 - 4 \)
\( = 27a + 36 + 5 = 27a + 41 \)
Calculate remainder \( R_2 = q(3) \):
\( q(3) = (3)^3 - 4(3) + a \)
\( = 27 - 12 + a = a + 15 \)
Since both remainders are equal, equate \( R_1 \) and \( R_2 \):
\( 27a + 41 = a + 15 \)
\( 27a - a = 15 - 41 \)
\( 26a = -26 \)
\( \implies a = -1 \).
Hence, the value of \( a \) is \( -1 \).
In simple words: Find p(3) and q(3), equate them since the remainders are identical, and solve for a.

Exam Tip: Don't forget to evaluate the constant terms on each side carefully: \( 36 + 9 - 4 = 41 \) and \( 27 - 12 = 15 \).

Question 10. Factorise by splitting the middle term:
i) 3x² - 14x + 8
ii) 4x² - 17x - 21
iii) 5x² - 16x - 21
iv) 6x² - x - 15
v) 7x² - 16x + 4
vi) 8x² - 6x - 9
Answer: To factorise \( ax^2 + bx + c \), find two numbers whose product is \( ac \) and whose sum is \( b \):
i) For \( 3x^2 - 14x + 8 \):
Product = \( 3 \times 8 = 24 \), Sum = \( -14 \). Numbers are \( -12 \) and \( -2 \).
\( 3x^2 - 12x - 2x + 8 = 3x(x - 4) - 2(x - 4) = (3x - 2)(x - 4) \).
ii) For \( 4x^2 - 17x - 21 \):
Product = \( 4 \times (-21) = -84 \), Sum = \( -17 \). Numbers are \( -21 \) and \( 4 \).
\( 4x^2 + 4x - 21x - 21 = 4x(x + 1) - 21(x + 1) = (4x - 21)(x + 1) \).
iii) For \( 5x^2 - 16x - 21 \):
Product = \( 5 \times (-21) = -105 \), Sum = \( -16 \). Numbers are \( -21 \) and \( 5 \).
\( 5x^2 + 5x - 21x - 21 = 5x(x + 1) - 21(x + 1) = (5x - 21)(x + 1) \).
iv) For \( 6x^2 - x - 15 \):
Product = \( 6 \times (-15) = -90 \), Sum = \( -1 \). Numbers are \( -10 \) and \( 9 \).
\( 6x^2 + 9x - 10x - 15 = 3x(2x + 3) - 5(2x + 3) = (3x - 5)(2x + 3) \).
v) For \( 7x^2 - 16x + 4 \):
Product = \( 7 \times 4 = 28 \), Sum = \( -16 \). Numbers are \( -14 \) and \( -2 \).
\( 7x^2 - 14x - 2x + 4 = 7x(x - 2) - 2(x - 2) = (7x - 2)(x - 2) \).
vi) For \( 8x^2 - 6x - 9 \):
Product = \( 8 \times (-9) = -72 \), Sum = \( -6 \). Numbers are \( -12 \) and \( 6 \).
\( 8x^2 - 12x + 6x - 9 = 4x(2x - 3) + 3(2x - 3) = (4x + 3)(2x - 3) \).
In simple words: Find two integers that multiply to give the product of the first and last coefficients and add up to the middle coefficient.

Exam Tip: Verify your factors by expanding them back to the original quadratic expression to catch simple sign mistakes.

Question 11. Factorise using factor theorem:
i) x³ + 6x² + 11x + 6
ii) x³ + 4x² + x - 6
iii) x³ - 13x - 12
iv) 2x³ + 3x² - 3x - 2
Answer: Find an integer root by trial, factor out the corresponding linear term, and factorise the remaining quadratic quotient:
i) Let \( p(x) = x^3 + 6x^2 + 11x + 6 \).
Trial value \( x = -1 \): \( p(-1) = (-1)^3 + 6(-1)^2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0 \).
\( \implies (x + 1) \) is a factor.
Dividing \( p(x) \) by \( x + 1 \):
\( x^3 + 6x^2 + 11x + 6 = (x + 1)(x^2 + 5x + 6) \)
Factorising \( x^2 + 5x + 6 = (x + 2)(x + 3) \).
Hence, \( x^3 + 6x^2 + 11x + 6 = (x + 1)(x + 2)(x + 3) \).

ii) Let \( p(x) = x^3 + 4x^2 + x - 6 \).
Trial value \( x = 1 \): \( p(1) = 1^3 + 4(1)^2 + 1 - 6 = 1 + 4 + 1 - 6 = 0 \).
\( \implies (x - 1) \) is a factor.
Dividing \( p(x) \) by \( x - 1 \):
\( x^3 + 4x^2 + x - 6 = (x - 1)(x^2 + 5x + 6) \)
Factorising \( x^2 + 5x + 6 = (x + 2)(x + 3) \).
Hence, \( x^3 + 4x^2 + x - 6 = (x - 1)(x + 2)(x + 3) \).

iii) Let \( p(x) = x^3 - 13x - 12 \).
Trial value \( x = -1 \): \( p(-1) = (-1)^3 - 13(-1) - 12 = -1 + 13 - 12 = 0 \).
\( \implies (x + 1) \) is a factor.
Dividing \( p(x) \) by \( x + 1 \):
\( x^3 - 13x - 12 = (x + 1)(x^2 - x - 12) \)
Factorising \( x^2 - x - 12 = (x - 4)(x + 3) \).
Hence, \( x^3 - 13x - 12 = (x + 1)(x + 3)(x - 4) \).

iv) Let \( p(x) = 2x^3 + 3x^2 - 3x - 2 \).
Trial value \( x = 1 \): \( p(1) = 2(1)^3 + 3(1)^2 - 3(1) - 2 = 2 + 3 - 3 - 2 = 0 \).
\( \implies (x - 1) \) is a factor.
Dividing \( p(x) \) by \( x - 1 \):
\( 2x^3 + 3x^2 - 3x - 2 = (x - 1)(2x^2 + 5x + 2) \)
Factorising \( 2x^2 + 5x + 2 = (2x + 1)(x + 2) \).
Hence, \( 2x^3 + 3x^2 - 3x - 2 = (x - 1)(x + 2)(2x + 1) \).
In simple words: Find one root by testing factors of the constant term, divide by the factor to get a quadratic, and factorise that quadratic.

Exam Tip: Try factors \( \pm 1, \pm 2 \) first as they are the most common roots in school examinations.

Question 12. Factorise using suitable identities:
i) 4x² - 36x + 81
ii) x² + 2 + 1/x²
iii) x³ - x
iv) x³ + 3x² + 3x + 1
v) 25x² - 9y²
vi) x³ + 1
vii) x³ - 3x + 3/x - 1/x³
viii) x³ - 8y³
ix) a³ + b³ - 1 + 3ab
x) a² + b² + 2ab + 2a + 2b + 1
Answer:
i) \( 4x^2 - 36x + 81 = (2x)^2 - 2(2x)(9) + 9^2 = (2x - 9)^2 \)
[Identity: \( a^2 - 2ab + b^2 = (a - b)^2 \)]

ii) \( x^2 + 2 + \frac{1}{x^2} = x^2 + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 \)
[Identity: \( a^2 + 2ab + b^2 = (a + b)^2 \)]

iii) \( x^3 - x = x(x^2 - 1) = x(x^2 - 1^2) = x(x - 1)(x + 1) \)
[Identity: \( a^2 - b^2 = (a - b)(a + b) \)]

iv) \( x^3 + 3x^2 + 3x + 1 = x^3 + 3(x)^2(1) + 3(x)(1)^2 + 1^3 = (x + 1)^3 \)
[Identity: \( a^3 + 3a^2b + 3ab^2 + b^3 = (a + b)^3 \)]

v) \( 25x^2 - 9y^2 = (5x)^2 - (3y)^2 = (5x - 3y)(5x + 3y) \)
[Identity: \( a^2 - b^2 = (a - b)(a + b) \)]

vi) \( x^3 + 1 = x^3 + 1^3 = (x + 1)(x^2 - x(1) + 1^2) = (x + 1)(x^2 - x + 1) \)
[Identity: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)]

vii) \( x^3 - 3x + \frac{3}{x} - \frac{1}{x^3} = x^3 - 3(x)^2\left(\frac{1}{x}\right) + 3(x)\left(\frac{1}{x}\right)^2 - \left(\frac{1}{x}\right)^3 = \left(x - \frac{1}{x}\right)^3 \)
[Identity: \( a^3 - 3a^2b + 3ab^2 - b^3 = (a - b)^3 \)]

viii) \( x^3 - 8y^3 = x^3 - (2y)^3 = (x - 2y)\left(x^2 + x(2y) + (2y)^2\right) = (x - 2y)(x^2 + 2xy + 4y^2) \)
[Identity: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)]

ix) \( a^3 + b^3 - 1 + 3ab = a^3 + b^3 + (-1)^3 - 3(a)(b)(-1) \)
\( = (a + b - 1)\left(a^2 + b^2 + (-1)^2 - ab - b(-1) - (-1)a\right) \)
\( = (a + b - 1)(a^2 + b^2 + 1 - ab + b + a) \)
[Identity: \( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \)]

x) \( a^2 + b^2 + 2ab + 2a + 2b + 1 = (a + b)^2 + 2(a + b)(1) + 1^2 = (a + b + 1)^2 \)
[Identity: \( X^2 + 2X + 1 = (X + 1)^2 \) where \( X = a + b \)]
In simple words: Match the algebraic pattern to a known formula such as difference of squares, cubes of binomials, or sum of cubes.

Exam Tip: Always state the standard algebraic identity used alongside each factorisation step to ensure full marks.

Question 13. Evaluate using suitable identities:
i) 103²
ii) 98 x 102
iii) 18 x 8
iv) (13)³
v) 51 x 49
vi) 97²
vii) 51² - 49²
viii) 103 x 102
ix) (-19)³ + 10³ + 9³
Answer:
i) \( 103^2 = (100 + 3)^2 = 100^2 + 2(100)(3) + 3^2 = 10000 + 600 + 9 = 10609 \)
[Identity: \( (a + b)^2 = a^2 + 2ab + b^2 \)]

ii) \( 98 \times 102 = (100 - 2)(100 + 2) = 100^2 - 2^2 = 10000 - 4 = 9996 \)
[Identity: \( (a - b)(a + b) = a^2 - b^2 \)]

iii) \( 18 \times 8 = (13 + 5)(13 - 5) = 13^2 - 5^2 = 169 - 25 = 144 \)
(Or using \( (10 + 8)(10 - 2) = 10^2 + (8 - 2)(10) + 8(-2) = 100 + 60 - 16 = 144 \))
[Identity: \( (a + b)(a - b) = a^2 - b^2 \)]

iv) \( (13)^3 = (10 + 3)^3 = 10^3 + 3^3 + 3(10)(3)(10 + 3) = 1000 + 27 + 90(13) = 1027 + 1170 = 2197 \)
[Identity: \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \)]

v) \( 51 \times 49 = (50 + 1)(50 - 1) = 50^2 - 1^2 = 2500 - 1 = 2499 \)
[Identity: \( (a + b)(a - b) = a^2 - b^2 \)]

vi) \( 97^2 = (100 - 3)^2 = 100^2 - 2(100)(3) + 3^2 = 10000 - 600 + 9 = 9409 \)
[Identity: \( (a - b)^2 = a^2 - 2ab + b^2 \)]

vii) \( 51^2 - 49^2 = (51 + 49)(51 - 49) = (100)(2) = 200 \)
[Identity: \( a^2 - b^2 = (a + b)(a - b) \)]

viii) \( 103 \times 102 = (100 + 3)(100 + 2) = 100^2 + (3 + 2)(100) + (3)(2) = 10000 + 500 + 6 = 10506 \)
[Identity: \( (x + a)(x + b) = x^2 + (a + b)x + ab \)]

ix) For \( (-19)^3 + 10^3 + 9^3 \):
Let \( x = -19 \), \( y = 10 \), and \( z = 9 \).
Check the sum: \( x + y + z = -19 + 10 + 9 = 0 \).
Since \( x + y + z = 0 \), the identity gives \( x^3 + y^3 + z^3 = 3xyz \):
\( (-19)^3 + 10^3 + 9^3 = 3(-19)(10)(9) = -57 \times 90 = -5130 \).
In simple words: Break numbers into convenient sums or differences close to powers of 10 to apply algebraic formulas.

Exam Tip: Never evaluate these by direct long multiplication; direct multiplication awards zero marks when the question asks to use identities.

Question 14. If x² + 25x + 84 = (x+a)(x+b) find the values of a and b.
Answer: Expanding the product on the right-hand side:
\( (x + a)(x + b) = x^2 + (a + b)x + ab \)
Equating this with \( x^2 + 25x + 84 \), compare coefficients:
\( a + b = 25 \)
\( ab = 84 \)
We look for two positive factors of 84 that add up to 25:
\( 21 \times 4 = 84 \) and \( 21 + 4 = 25 \).
Therefore, \( a = 21 \) and \( b = 4 \) (or \( a = 4 \) and \( b = 21 \)).
In simple words: Find two numbers whose sum is 25 and whose product is 84, which gives 21 and 4.

Exam Tip: State that \( a \) and \( b \) are interchangeable because multiplication of factors is commutative.

Question 15. If x² + 5√5x + 30 = (x+a)(x+3√5) find a.
Answer: Expand the right side:
\( (x + a)(x + 3\sqrt{5}) = x^2 + (a + 3\sqrt{5})x + 3\sqrt{5}a \)
Comparing coefficients with \( x^2 + 5\sqrt{5}x + 30 \):
Matching the coefficient of \( x \):
\( a + 3\sqrt{5} = 5\sqrt{5} \)
\( \implies a = 5\sqrt{5} - 3\sqrt{5} = 2\sqrt{5} \)
Verification using the constant term:
\( 3\sqrt{5} \times a = 3\sqrt{5} \times 2\sqrt{5} = 6 \times 5 = 30 \), which matches the constant term.
Thus, the value of \( a \) is \( 2\sqrt{5} \).
In simple words: Equate the coefficient of x on both sides to find a, then verify with the constant term.

Exam Tip: Always verify your result with the constant term to ensure consistency between the radical terms.

Question 16. Factorise x² - 8x + 15 by using Factor Theorem
Answer: Let \( p(x) = x^2 - 8x + 15 \).
The constant term is 15. Its integer factors are \( \pm 1, \pm 3, \pm 5, \pm 15 \).
Testing \( x = 3 \):
\( p(3) = (3)^2 - 8(3) + 15 = 9 - 24 + 15 = 0 \)
Since \( p(3) = 0 \), by the Factor Theorem, \( (x - 3) \) is a factor of \( p(x) \).
Testing \( x = 5 \):
\( p(5) = (5)^2 - 8(5) + 15 = 25 - 40 + 15 = 0 \)
Since \( p(5) = 0 \), by the Factor Theorem, \( (x - 5) \) is also a factor of \( p(x) \).
Because \( p(x) \) is of degree 2, it can have at most two linear factors:
\( x^2 - 8x + 15 = (x - 3)(x - 5) \).
In simple words: Find the roots 3 and 5 by testing factors of 15, then use the Factor Theorem to write the factors.

Exam Tip: State that finding \( p(c) = 0 \) implies \( (x - c) \) is a factor under the Factor Theorem.

Question 17. If 2x³ + 5x² + 5x + k is exactly divisible by x² + x + 1 find the value of k
Answer: Divide \( 2x^3 + 5x^2 + 5x + k \) by \( x^2 + x + 1 \) using polynomial division:
1. Divide the leading term: \( \frac{2x^3}{x^2} = 2x \).
Multiply divisor: \( 2x(x^2 + x + 1) = 2x^3 + 2x^2 + 2x \).
Subtract:
\( (2x^3 + 5x^2 + 5x + k) - (2x^3 + 2x^2 + 2x) = 3x^2 + 3x + k \).
2. Divide the new leading term: \( \frac{3x^2}{x^2} = 3 \).
Multiply divisor: \( 3(x^2 + x + 1) = 3x^2 + 3x + 3 \).
Subtract:
\( (3x^2 + 3x + k) - (3x^2 + 3x + 3) = k - 3 \).
For exact divisibility, the remainder must equal 0:
\( k - 3 = 0 \)
\( \implies k = 3 \).
Hence, the value of \( k \) is 3.
In simple words: Perform polynomial long division to find the remainder k - 3, then set it equal to 0.

Exam Tip: Set the final remainder expression equal to zero to solve for the unknown constant.