CBSE Class 9 Mathematics Polynomials Worksheet Set 04

Very Short Answer Type Questions

Question. How many terms are there in the polynomial 5 – 3t + 2t⁶ ?
Answer : Number of terms = 3

Question. If, p (− 1/4) = 0, where p(x) = 8x³ – ax² – x + 2, then find the value of a.
Answer : ∵ p(−1/4) = 0 
⇒ 8(-1/4)³ - a (- 1/4)² - (- 1/4) + 2 = 0 
⇒ -8/64 - a/16 + 1/4 + 2 = 0
⇒ -1/8 + 1/4 + 2 = a/16 ⇒ -1+2+16/8 = a/16
⇒ a = 17 × 2
∴ a = 34

Question. Find the product of (lx +my) and (lx – my).
Answer : Product of (lx + my) and (lx – my) = (lx + my) (lx – my)
= (lx)² – (my)² = l²x² – m²y²

Question. Find the value of p(x) = √2x² + √2x + 6 at x = √2
Answer : p (√2) = √2(√2)² + √2 (√2) + 6
=2√2 + 2 + 6
=2√2 + 8

Question. Find the factors of x² + 2x – 15.
Answer : We have, x² + 2x – 15 = x² + 5x – 3x – 15
= x(x + 5) – 3(x + 5) = (x + 5) (x – 3)
∴ Factors of x² + 2x – 15 are (x + 5) and (x – 3).

Question. Find the zeroes of the polynomial 3x + p.
Answer : 3x + π = 0 ⇒ 3x = – π ⇒ x = − π/3
∴ x = − π/3 is the zero of the polynomial 3x + p.

Question. Factorise : a3 – b3 + 3ba² – 3ab²
Answer : We have, a³ – b3 + 3a²b – 3ab²
= (a – b) (a² + b² + ab) + 3ab (a – b)
= (a – b) [a² + b² + 4ab]

Question. Factorise : 14a2b – 7ab²
Answer : We have, 14a²b – 7ab² = 7ab(2a – b)

Question. Find the zero of the polynomial q(x) = 3x + 2.
Answer : We have, q(x) = 3x + 2
Put q(x) = 0 ⇒ 3x + 2 = 0 ⇒ x = −2/3
∴ −2/3 is the zero of the polynomial q(x) = 3x + 2.

Question. Is x + 2 a zero of p(x) = 2x³ – x² + 3?
Answer : (x + 2) will be a zero of p(x), if p (–2) = 0
∴ p(–2) = 2(–2)³ – (–2)² + 3 = –16 – 4 + 3 = –17 ≠ 0
∴ x + 2 is not a zero of p(x).

Question 1. Factorize the following by splitting the middle term:
(a) \( 3x^2 + 19x + 30 \)
(b) \( 2\sqrt{2}x^2 + 9x + 5\sqrt{2} \)
(c) \( 4x^2 - 13x + 10 \)
Answer:
(a) We need two numbers whose product is \( 3 \times 30 = 90 \) and whose sum is \( 19 \). The numbers are \( 9 \) and \( 10 \).
\( 3x^2 + 19x + 30 = 3x^2 + 9x + 10x + 30 \)
\( = 3x(x + 3) + 10(x + 3) \)
\( = (x + 3)(3x + 10) \)

(b) The product required is \( 2\sqrt{2} \times 5\sqrt{2} = 20 \), and the sum is \( 9 \). The numbers are \( 4 \) and \( 5 \).
\( 2\sqrt{2}x^2 + 9x + 5\sqrt{2} = 2\sqrt{2}x^2 + 4x + 5x + 5\sqrt{2} \)
\( = 2\sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) \)
\( = (x + \sqrt{2})(2\sqrt{2}x + 5) \)

(c) Here, the product is \( 4 \times 10 = 40 \) and the sum is \( -13 \). The numbers are \( -8 \) and \( -5 \).
\( 4x^2 - 13x + 10 = 4x^2 - 8x - 5x + 10 \)
\( = 4x(x - 2) - 5(x - 2) \)
\( = (x - 2)(4x - 5) \)
In simple words: Find two integers that multiply to give the product of the first and last coefficients while adding up to the middle coefficient, then group terms to factor out common binomials.

Exam Tip: When dealing with square root coefficients, multiply the radical terms carefully to find the constant product before splitting the middle term.

Question 2. Factorize the following by Factor theorem:
(a) \( x^3 + 9x^2 + 23x + 15 \)
(b) \( x^3 + 6x^2 + 11x + 6 \)
Answer:
(a) Let \( P(x) = x^3 + 9x^2 + 23x + 15 \). The factors of the constant term \( 15 \) are \( \pm 1, \pm 3, \pm 5, \pm 15 \).
Testing \( x = -1 \):
\( P(-1) = (-1)^3 + 9(-1)^2 + 23(-1) + 15 = -1 + 9 - 23 + 15 = 0 \)
Since \( P(-1) = 0 \), \( (x + 1) \) is a factor.
Dividing \( P(x) \) by \( (x + 1) \):
\( x^3 + 9x^2 + 23x + 15 = x^2(x + 1) + 8x(x + 1) + 15(x + 1) \)
\( = (x + 1)(x^2 + 8x + 15) \)
Factoring the quadratic part:
\( x^2 + 8x + 15 = (x + 3)(x + 5) \)
Thus, \( x^3 + 9x^2 + 23x + 15 = (x + 1)(x + 3)(x + 5) \).

(b) Let \( P(x) = x^3 + 6x^2 + 11x + 6 \). The factors of \( 6 \) are \( \pm 1, \pm 2, \pm 3, \pm 6 \).
Testing \( x = -1 \):
\( P(-1) = (-1)^3 + 6(-1)^2 + 11(-1) + 6 = -1 + 6 - 11 + 6 = 0 \)
Therefore, \( (x + 1) \) is a factor.
Dividing \( P(x) \) by \( (x + 1) \):
\( x^3 + 6x^2 + 11x + 6 = x^2(x + 1) + 5x(x + 1) + 6(x + 1) \)
\( = (x + 1)(x^2 + 5x + 6) \)
Factoring the quadratic factor:
\( x^2 + 5x + 6 = (x + 2)(x + 3) \)
Hence, \( x^3 + 6x^2 + 11x + 6 = (x + 1)(x + 2)(x + 3) \).
In simple words: Test integer factors of the constant term to find one root that makes the polynomial zero, divide by the corresponding linear factor, and then factor the resulting quadratic expression.

Exam Tip: If all terms in a polynomial have positive signs, only test negative factors of the constant term because positive substitutions will never equal zero.

Question 3. Factorize the following by using a suitable identity:
(a) \( 4x^2 + 12xy + 9y^2 \)
(b) \( 2a^5 - 54a^2 \)
(c) \( 2\sqrt{2}x^3 + 3\sqrt{3}y^3 \)
(d) \( x^5 - x \)
(e) \( x^6 - y^6 \)
(f) \( (a - b)^3 + (b - c)^3 + (c - a)^3 \)
(g) \( x^8 - y^8 \)
(h) \( 27x^3 - 135x^2 + 225x - 125 \)
Answer:
(a) Using \( A^2 + 2AB + B^2 = (A + B)^2 \):
\( 4x^2 + 12xy + 9y^2 = (2x)^2 + 2(2x)(3y) + (3y)^2 = (2x + 3y)^2 \)

(b) Factoring out the common term \( 2a^2 \) first:
\( 2a^5 - 54a^2 = 2a^2(a^3 - 27) = 2a^2(a^3 - 3^3) \)
Using \( A^3 - B^3 = (A - B)(A^2 + AB + B^2) \):
\( = 2a^2(a - 3)(a^2 + 3a + 9) \)

(c) Rewriting with cube terms:
\( 2\sqrt{2}x^3 + 3\sqrt{3}y^3 = (\sqrt{2}x)^3 + (\sqrt{3}y)^3 \)
Applying \( A^3 + B^3 = (A + B)(A^2 - AB + B^2) \):
\( = (\sqrt{2}x + \sqrt{3}y)[(\sqrt{2}x)^2 - (\sqrt{2}x)(\sqrt{3}y) + (\sqrt{3}y)^2] \)
\( = (\sqrt{2}x + \sqrt{3}y)(2x^2 - \sqrt{6}xy + 3y^2) \)

(d) Factoring out \( x \):
\( x^5 - x = x(x^4 - 1) = x[(x^2)^2 - 1^2] \)
\( = x(x^2 - 1)(x^2 + 1) = x(x - 1)(x + 1)(x^2 + 1) \)

(e) Using difference of squares first:
\( x^6 - y^6 = (x^3)^2 - (y^3)^2 = (x^3 - y^3)(x^3 + y^3) \)
Applying sum and difference of cubes identities:
\( = (x - y)(x^2 + xy + y^2)(x + y)(x^2 - xy + y^2) \)
\( = (x - y)(x + y)(x^2 + xy + y^2)(x^2 - xy + y^2) \)

(f) Let \( x = a - b \), \( y = b - c \), and \( z = c - a \).
Since \( x + y + z = (a - b) + (b - c) + (c - a) = 0 \), we use the identity \( x^3 + y^3 + z^3 = 3xyz \):
\( (a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a) \)

(g) Repeatedly applying difference of squares:
\( x^8 - y^8 = (x^4)^2 - (y^4)^2 = (x^4 - y^4)(x^4 + y^4) \)
\( = (x^2 - y^2)(x^2 + y^2)(x^4 + y^4) \)
\( = (x - y)(x + y)(x^2 + y^2)(x^4 + y^4) \)

(h) Expressing in the form of \( A^3 - 3A^2B + 3AB^2 - B^3 = (A - B)^3 \):
\( 27x^3 - 135x^2 + 225x - 125 = (3x)^3 - 3(3x)^2(5) + 3(3x)(5^2) - 5^3 \)
\( = (3x - 5)^3 \)
In simple words: Match each expression to standard algebraic patterns such as perfect squares, sum or difference of cubes, or the conditional identity where variables summing to zero yield three times their product.

Exam Tip: For expressions like \( x^6 - y^6 \), always apply the difference of squares before the difference of cubes to achieve complete factorization without complicated intermediate steps.

Question 4. Evaluate the following using a suitable identity:
(a) \( 998^3 \)
(b) \( 10.2^3 \)
(c) \( 998^2 - 4 \)
(d) \( 999^2 - 1 \)
(e) \( (-25)^3 + 10^3 + 15^3 \)
(f) \( 10.2 \times 9.8 \)
Answer:
(a) Using \( (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \) with \( a = 1000 \) and \( b = 2 \):
\( 998^3 = (1000 - 2)^3 \)
\( = 1000^3 - 3(1000)^2(2) + 3(1000)(2^2) - 2^3 \)
\( = 1,000,000,000 - 6,000,000 + 12,000 - 8 \)
\( = 994,000,000 + 11,992 = 994,011,992 \)

(b) Using \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) with \( a = 10 \) and \( b = 0.2 \):
\( 10.2^3 = (10 + 0.2)^3 \)
\( = 10^3 + 3(10^2)(0.2) + 3(10)(0.2^2) + 0.2^3 \)
\( = 1000 + 60 + 1.2 + 0.008 = 1061.208 \)

(c) Using \( a^2 - b^2 = (a - b)(a + b) \) with \( a = 998 \) and \( b = 2 \):
\( 998^2 - 4 = 998^2 - 2^2 \)
\( = (998 - 2)(998 + 2) = (996)(1000) = 996,000 \)

(d) Using \( a^2 - b^2 = (a - b)(a + b) \) with \( a = 999 \) and \( b = 1 \):
\( 999^2 - 1 = 999^2 - 1^2 \)
\( = (999 - 1)(999 + 1) = (998)(1000) = 998,000 \)

(e) Let \( a = -25 \), \( b = 10 \), and \( c = 15 \).
Check their sum: \( a + b + c = -25 + 10 + 15 = 0 \).
When \( a + b + c = 0 \), \( a^3 + b^3 + c^3 = 3abc \):
\( (-25)^3 + 10^3 + 15^3 = 3(-25)(10)(15) = -11,250 \)

(f) Using \( (a + b)(a - b) = a^2 - b^2 \) with \( a = 10 \) and \( b = 0.2 \):
\( 10.2 \times 9.8 = (10 + 0.2)(10 - 0.2) \)
\( = 10^2 - (0.2)^2 = 100 - 0.04 = 99.96 \)
In simple words: Write numbers around convenient base values like 10 or 1000 so that algebraic identities calculate large powers and products without direct multiplication.

Exam Tip: Never compute these products by long multiplication; examiners award marks strictly for naming and substituting into the appropriate algebraic identity.

Question 5. If 5 is a zero of \( x^3 + kx^2 + 2x + 8 \), find k.
Answer: Let \( P(x) = x^3 + kx^2 + 2x + 8 \).
Because 5 is a zero of the polynomial, \( P(5) = 0 \).
\( 5^3 + k(5^2) + 2(5) + 8 = 0 \)
\( 125 + 25k + 10 + 8 = 0 \)
\( 25k + 143 = 0 \)
\( 25k = -143 \)

\( \implies k = -\frac{143}{25} = -5.72 \)
In simple words: Substitute x with 5, equate the entire expression to zero, and solve the linear equation for k.

Exam Tip: A zero of a polynomial \( P(x) \) simply means that replacing \( x \) with that number produces a result of zero.

Question 6. If \( (x - 2) \) is a zero of \( x^3 - 4x^2 + kx - 8 \), find k.
Answer: If \( (x - 2) \) is a factor (meaning \( x = 2 \) is a zero of the polynomial):
Let \( P(x) = x^3 - 4x^2 + kx - 8 \).
Setting \( P(2) = 0 \):
\( 2^3 - 4(2^2) + k(2) - 8 = 0 \)
\( 8 - 16 + 2k - 8 = 0 \)
\( 2k - 16 = 0 \)
\( 2k = 16 \)

\( \implies k = 8 \)
In simple words: Setting the factor equal to zero gives x = 2; substituting this into the polynomial makes it vanish, allowing us to find k = 8.

Exam Tip: Simplify the constant numbers first before isolating the term containing the unknown variable.

Question 7. If \( (x - 2) \) and \( (x + 3) \) are factors of \( x^3 + ax^2 + bx - 30 \), find a and b.
Answer: Let \( P(x) = x^3 + ax^2 + bx - 30 \).
By the Factor Theorem, since \( (x - 2) \) is a factor, \( P(2) = 0 \):
\( 2^3 + a(2^2) + b(2) - 30 = 0 \)
\( 8 + 4a + 2b - 30 = 0 \)
\( 4a + 2b = 22 \)
Dividing by 2:
\( 2a + b = 11 \quad \text{--- (1)} \)

Similarly, because \( (x + 3) \) is a factor, \( P(-3) = 0 \):
\( (-3)^3 + a(-3)^2 + b(-3) - 30 = 0 \)
\( -27 + 9a - 3b - 30 = 0 \)
\( 9a - 3b = 57 \)
Dividing by 3:
\( 3a - b = 19 \quad \text{--- (2)} \)

Adding equations (1) and (2):
\( (2a + b) + (3a - b) = 11 + 19 \)
\( 5a = 30 \implies a = 6 \)
Substituting \( a = 6 \) into equation (1):
\( 2(6) + b = 11 \implies 12 + b = 11 \implies b = -1 \)
Therefore, \( a = 6 \) and \( b = -1 \).
In simple words: Set up two linear equations by evaluating the polynomial at 2 and -3, then solve them simultaneously to find the values of a and b.

Exam Tip: Be careful with signs when substituting negative roots like \( x = -3 \); note that \( (-3)^3 = -27 \) while \( (-3)^2 = +9 \).

Question 8. If \( x + y + z = 8 \) and \( xy + yz + zx = 20 \), find the value of \( x^3 + y^3 + z^3 - 3xyz \).
Answer: We first find \( x^2 + y^2 + z^2 \) using the expansion of \( (x + y + z)^2 \):
\( (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \)
\( 8^2 = x^2 + y^2 + z^2 + 2(20) \)
\( 64 = x^2 + y^2 + z^2 + 40 \)

\( \implies x^2 + y^2 + z^2 = 64 - 40 = 24 \)
Now, we evaluate using the identity:
\( x^3 + y^3 + z^3 - 3xyz = (x + y + z)[x^2 + y^2 + z^2 - (xy + yz + zx)] \)
\( = 8 \times (24 - 20) \)
\( = 8 \times 4 = 32 \)
In simple words: First determine the sum of the squares by expanding the square of the sum, then plug that value into the cubic identity.

Exam Tip: Remember to subtract the sum of pairwise products inside the second factor: \( x^2 + y^2 + z^2 - (xy + yz + zx) \).

Question 9. If \( a + b + c = 9 \) and \( a^2 + b^2 + c^2 = 35 \), find the value of \( a^3 + b^3 + c^3 - 3abc \).
Answer: First, find \( ab + bc + ca \) using the identity for \( (a + b + c)^2 \):
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \)
\( 9^2 = 35 + 2(ab + bc + ca) \)
\( 81 - 35 = 2(ab + bc + ca) \)
\( 46 = 2(ab + bc + ca) \implies ab + bc + ca = 23 \)
Now, apply the cubic identity:
\( a^3 + b^3 + c^3 - 3abc = (a + b + c)[a^2 + b^2 + c^2 - (ab + bc + ca)] \)
\( = 9 \times (35 - 23) \)
\( = 9 \times 12 = 108 \)
In simple words: Compute the missing pairwise product term from the squared sum, then multiply the sum by the difference of squares and pairwise products.

Exam Tip: State the identity clearly before substituting the numerical values to guarantee step marks.

Question 10. Find the value of \( 2.7^3 - 1.6^3 - 1.1^3 \) using a suitable identity.
Answer: We can rewrite the expression as:
\( 2.7^3 + (-1.6)^3 + (-1.1)^3 \)
Let \( x = 2.7 \), \( y = -1.6 \), and \( z = -1.1 \).
Check their sum:
\( x + y + z = 2.7 + (-1.6) + (-1.1) = 2.7 - 2.7 = 0 \)
Using the identity that if \( x + y + z = 0 \), then \( x^3 + y^3 + z^3 = 3xyz \):
\( 2.7^3 + (-1.6)^3 + (-1.1)^3 = 3(2.7)(-1.6)(-1.1) \)
\( = 3 \times 2.7 \times 1.76 \)
\( = 8.1 \times 1.76 = 14.256 \)
In simple words: Because the three base numbers add up to zero, their cubes simply equal three times their product.

Exam Tip: Always show that the sum of the terms is zero before applying \( x^3 + y^3 + z^3 = 3xyz \).

Question 11. Factorize the following by using a suitable identity:
(a) \( a^3 + b^3 - 8c^3 + 6abc \)
(b) \( (a/b)^3 + (b/c)^3 + (c/a)^3 - 3 \)
(c) \( 8x^3 - 27y^3 + 125z^3 + 90xyz \)
Answer:
(a) Rewriting the expression:
\( a^3 + b^3 + (-2c)^3 - 3(a)(b)(-2c) \)
Using the identity \( x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \):
\( = [a + b + (-2c)][a^2 + b^2 + (-2c)^2 - ab - b(-2c) - (-2c)a] \)
\( = (a + b - 2c)(a^2 + b^2 + 4c^2 - ab + 2bc + 2ca) \)

(b) Let \( x = \frac{a}{b} \), \( y = \frac{b}{c} \), and \( z = \frac{c}{a} \).
Notice that \( xyz = \left(\frac{a}{b}\right)\left(\frac{b}{c}\right)\left(\frac{c}{a}\right) = 1 \), so \( 3 = 3xyz \).
The expression becomes \( x^3 + y^3 + z^3 - 3xyz \):
\( = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) \)
Since \( xy = \frac{a}{c} \), \( yz = \frac{b}{a} \), and \( zx = \frac{c}{b} \), we have:
\( = \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right)\left(\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} - \frac{a}{c} - \frac{b}{a} - \frac{c}{b}\right) \)

(c) Expressing each term as a cube:
\( (2x)^3 + (-3y)^3 + (5z)^3 - 3(2x)(-3y)(5z) \)
Using the same identity with \( A = 2x \), \( B = -3y \), and \( C = 5z \):
\( = [2x + (-3y) + 5z][(2x)^2 + (-3y)^2 + (5z)^2 - (2x)(-3y) - (-3y)(5z) - (5z)(2x)] \)
\( = (2x - 3y + 5z)(4x^2 + 9y^2 + 25z^2 + 6xy + 15yz - 10zx) \)
In simple words: Represent each term as a cube so that the expression matches the identity for the sum of three cubes minus three times their product.

Exam Tip: Watch out for negative signs inside cubes; for example, \( -8c^3 = (-2c)^3 \) makes the third variable negative in the identity.

Question 12. Find the value of \( a^3 + 8b^3 \), if \( a + 2b = 10 \) and \( ab = 15 \).
Answer: We use the identity for the cube of a binomial:
\( (a + 2b)^3 = a^3 + (2b)^3 + 3(a)(2b)(a + 2b) \)
\( (a + 2b)^3 = a^3 + 8b^3 + 6ab(a + 2b) \)
Substituting \( a + 2b = 10 \) and \( ab = 15 \):
\( 10^3 = a^3 + 8b^3 + 6(15)(10) \)
\( 1000 = a^3 + 8b^3 + 900 \)
\( a^3 + 8b^3 = 1000 - 900 \)

\( \implies a^3 + 8b^3 = 100 \)
In simple words: Expand the cube of the given linear expression and substitute the known numbers to solve for the sum of the cubes.

Exam Tip: Be sure to multiply the cross-term properly: \( 3 \times a \times 2b = 6ab \), not \( 3ab \).

Question 13. Find the value of \( a^3 + 27b^3 \), if \( a - 2b = (-6) \) and \( ab = (-10) \).
Answer: Using the standard algebraic relation with \( a - 3b = -6 \) and \( ab = -10 \):
\( (a - 3b)^3 = a^3 - 27b^3 - 9ab(a - 3b) \)
Substituting the values:
\( (-6)^3 = a^3 - 27b^3 - 9(-10)(-6) \)
\( -216 = a^3 - 27b^3 - 540 \)
\( a^3 - 27b^3 = 540 - 216 = 324 \)

Alternatively, with \( a + 3b = -6 \) and \( ab = -10 \):
\( (a + 3b)^3 = a^3 + 27b^3 + 9ab(a + 3b) \)
\( (-6)^3 = a^3 + 27b^3 + 9(-10)(-6) \)
\( -216 = a^3 + 27b^3 + 540 \)

\( \implies a^3 + 27b^3 = -216 - 540 = -756 \)
In simple words: Cube the given linear relation and substitute the product of the terms to isolate the cubic expression.

Exam Tip: Check whether a sum or difference of cubes is requested to select the correct sign in the binomial expansion.

Question 14. Find the value of m and n if \( y^2 - 1 \) is a factor of \( y^4 + my^3 + 2y^2 - 3y + n \).
Answer: Factoring the divisor gives \( y^2 - 1 = (y - 1)(y + 1) \).
Since \( y^2 - 1 \) is a factor of \( P(y) = y^4 + my^3 + 2y^2 - 3y + n \), both \( (y - 1) \) and \( (y + 1) \) must be factors, meaning \( P(1) = 0 \) and \( P(-1) = 0 \).

1. For \( y = 1 \):
\( P(1) = 1^4 + m(1^3) + 2(1^2) - 3(1) + n = 0 \)
\( 1 + m + 2 - 3 + n = 0 \)
\( m + n = 0 \quad \text{--- (1)} \)

2. For \( y = -1 \):
\( P(-1) = (-1)^4 + m(-1)^3 + 2(-1)^2 - 3(-1) + n = 0 \)
\( 1 - m + 2 + 3 + n = 0 \)
\( -m + n + 6 = 0 \implies -m + n = -6 \quad \text{--- (2)} \)

Adding equations (1) and (2):
\( (m + n) + (-m + n) = 0 + (-6) \)
\( 2n = -6 \implies n = -3 \)
From equation (1), \( m = -n = -(-3) = 3 \).
Therefore, \( m = 3 \) and \( n = -3 \).
In simple words: Factor the quadratic divisor into two linear terms, evaluate the polynomial at 1 and -1 to get two simple equations, and solve for m and n.

Exam Tip: When a quadratic factor factors as \( (y - a)(y + a) \), both \( a \) and \( -a \) must independently give a remainder of zero.

Question 15. \( (x + 2) \) is a factor of \( mx^3 + nx^2 + x - 6 \). It leaves the remainder 4 when divided by \( (x - 2) \). Find m and n.
Answer: Let \( P(x) = mx^3 + nx^2 + x - 6 \).
Since \( (x + 2) \) is a factor, \( P(-2) = 0 \):
\( m(-2)^3 + n(-2)^2 + (-2) - 6 = 0 \)
\( -8m + 4n - 8 = 0 \)
Dividing by 4:
\( -2m + n = 2 \quad \text{--- (1)} \)

When divided by \( (x - 2) \), the remainder is 4, so \( P(2) = 4 \):
\( m(2)^3 + n(2)^2 + 2 - 6 = 4 \)
\( 8m + 4n - 4 = 4 \)
\( 8m + 4n = 8 \)
Dividing by 4:
\( 2m + n = 2 \quad \text{--- (2)} \)

Adding equations (1) and (2):
\( (-2m + n) + (2m + n) = 2 + 2 \)
\( 2n = 4 \implies n = 2 \)
Substituting \( n = 2 \) into equation (2):
\( 2m + 2 = 2 \implies 2m = 0 \implies m = 0 \)
Hence, \( m = 0 \) and \( n = 2 \).
In simple words: Use the Factor Theorem for x = -2 and the Remainder Theorem for x = 2 to set up two equations that yield m = 0 and n = 2.

Exam Tip: Be careful to set \( P(2) \) equal to the stated remainder 4, not 0.

Question 16. The polynomials \( kx^3 + 3x^2 - 3 \) and \( 2x^3 - 5x + k \) leave the same remainder when divided by \( (x - 4) \). Find k.
Answer: Let \( P(x) = kx^3 + 3x^2 - 3 \) and \( Q(x) = 2x^3 - 5x + k \).
By the Remainder Theorem, the remainders when divided by \( (x - 4) \) are \( P(4) \) and \( Q(4) \).
Evaluating \( P(4) \):
\( P(4) = k(4^3) + 3(4^2) - 3 = 64k + 3(16) - 3 = 64k + 45 \)
Evaluating \( Q(4) \):
\( Q(4) = 2(4^3) - 5(4) + k = 2(64) - 20 + k = 128 - 20 + k = 108 + k \)
Since the remainders are equal, \( P(4) = Q(4) \):
\( 64k + 45 = 108 + k \)
\( 64k - k = 108 - 45 \)
\( 63k = 63 \)

\( \implies k = 1 \)
In simple words: Substitute x = 4 into both polynomials, set the two expressions equal to each other, and solve for k.

Exam Tip: Keep powers of 4 organized (\( 4^3 = 64 \), \( 4^2 = 16 \)) to prevent arithmetic errors when equating the remainders.

Question 17. Factorize: \( x^3 + \frac{1}{x^3} - 2 \).
Answer: Writing the expression with a common denominator of \( x^3 \):
\( x^3 + \frac{1}{x^3} - 2 = \frac{x^6 - 2x^3 + 1}{x^3} \)
The numerator is a perfect square of the form \( (A - 1)^2 \) where \( A = x^3 \):
\( x^6 - 2x^3 + 1 = (x^3 - 1)^2 \)
Now, factoring \( x^3 - 1 \) using the difference of cubes identity \( A^3 - B^3 = (A - B)(A^2 + AB + B^2) \):
\( x^3 - 1 = (x - 1)(x^2 + x + 1) \)
Therefore, squaring this product gives:
\( (x^3 - 1)^2 = (x - 1)^2(x^2 + x + 1)^2 \)
Substituting back over \( x^3 \):
\( x^3 + \frac{1}{x^3} - 2 = \frac{1}{x^3}(x - 1)^2(x^2 + x + 1)^2 \)
In simple words: Combine terms into a single fraction to reveal a perfect square in the numerator, then factor the cubic term using the difference of cubes.

Exam Tip: Recognize \( u + \frac{1}{u} - 2 \) as \( \left(\sqrt{u} - \frac{1}{\sqrt{u}}\right)^2 \) or combine over a common denominator to spot the hidden perfect square.

Question 18. If \( ab + bc + ca = 10 \) and \( a^2 + b^2 + c^2 = 44 \), find \( a^3 + b^3 + c^3 - 3abc \).
Answer: First, determine \( a + b + c \) using the algebraic expansion:
\( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \)
\( (a + b + c)^2 = 44 + 2(10) = 44 + 20 = 64 \)
Taking the square root:
\( a + b + c = \pm 8 \)
Now, use the cubic identity:
\( a^3 + b^3 + c^3 - 3abc = (a + b + c)[a^2 + b^2 + c^2 - (ab + bc + ca)] \)
\( = (a + b + c)(44 - 10) = 34(a + b + c) \)
When \( a + b + c = 8 \):
\( a^3 + b^3 + c^3 - 3abc = 34 \times 8 = 272 \)
When \( a + b + c = -8 \):
\( a^3 + b^3 + c^3 - 3abc = 34 \times (-8) = -272 \)
Thus, the value is \( \pm 272 \).
In simple words: Find the sum of the variables by taking the square root of 64, then multiply by the difference between the sum of squares and pairwise products.

Exam Tip: Mention both positive and negative values \( \pm 272 \) because the square root can take either sign unless variables are specified to be positive.

Question 19. Show that \( x - 2 \) is a factor \( p(x) = x^3 - 12x^2 + 44x - 48 \).
Answer: By the Factor Theorem, \( (x - 2) \) is a factor of \( p(x) \) if and only if \( p(2) = 0 \).
Substituting \( x = 2 \) into \( p(x) \):
\( p(2) = 2^3 - 12(2^2) + 44(2) - 48 \)
\( = 8 - 12(4) + 88 - 48 \)
\( = 8 - 48 + 88 - 48 \)
\( = (8 + 88) - (48 + 48) \)
\( = 96 - 96 = 0 \)
Since the remainder \( p(2) = 0 \), \( (x - 2) \) is confirmed to be a factor of \( p(x) \).
In simple words: Plug 2 into the polynomial, calculate the arithmetic value, and verify that the result equals zero.

Exam Tip: Group positive and negative terms separately to calculate the sum accurately and demonstrate that they cancel to zero.

Question 20. Factorize \( x^3 - 2x^2 - 5x + 6 \).
Answer: Let \( P(x) = x^3 - 2x^2 - 5x + 6 \). The factors of the constant term 6 are \( \pm 1, \pm 2, \pm 3, \pm 6 \).
Testing \( x = 1 \):
\( P(1) = 1^3 - 2(1^2) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0 \)
Since \( P(1) = 0 \), \( (x - 1) \) is a factor.
Dividing \( P(x) \) by \( (x - 1) \):
\( x^3 - 2x^2 - 5x + 6 = x^2(x - 1) - x(x - 1) - 6(x - 1) \)
\( = (x - 1)(x^2 - x - 6) \)
Factoring the quadratic quotient by middle term splitting:
\( x^2 - x - 6 = x^2 - 3x + 2x - 6 = x(x - 3) + 2(x - 3) = (x - 3)(x + 2) \)
Hence, \( x^3 - 2x^2 - 5x + 6 = (x - 1)(x - 3)(x + 2) \).
In simple words: Substitute x = 1 to find the first factor (x - 1), then factor the remaining quadratic quotient into (x - 3) and (x + 2).

Exam Tip: Always write the final answer as the product of all three linear factors together.

Question 21. If \( x + p \) is a factor of \( p(x) = x^5 - p^2 x^3 + 2x + p + 1 \), find p.
Answer: By the Factor Theorem, if \( (x + p) \) is a factor of \( p(x) \), then \( p(-p) = 0 \).
Substituting \( x = -p \) into the polynomial:
\( (-p)^5 - p^2(-p)^3 + 2(-p) + p + 1 = 0 \)
\( -p^5 - p^2(-p^3) - 2p + p + 1 = 0 \)
\( -p^5 + p^5 - p + 1 = 0 \)
\( -p + 1 = 0 \)

\( \implies p = 1 \)
In simple words: Substitute -p into the polynomial to eliminate the fifth-power terms, leaving a simple linear equation that gives p = 1.

Exam Tip: Take care with the sign of odd powers: \( (-p)^5 = -p^5 \) and \( (-p)^3 = -p^3 \), which causes \( -p^2(-p^3) \) to become \( +p^5 \).

Question 22. Factorize:
(a) \( a^2 + \frac{1}{a^2} + 2 \)
(b) \( 3(x + 2)^2 + 17(x + 2) + 10 \)
(c) \( 4x^2 - 9y^2 + 20x + 25 \)
Answer:
(a) Notice that \( 2 = 2 \cdot a \cdot \frac{1}{a} \). Using \( A^2 + 2AB + B^2 = (A + B)^2 \):
\( a^2 + \frac{1}{a^2} + 2 = a^2 + 2(a)\left(\frac{1}{a}\right) + \left(\frac{1}{a}\right)^2 \)
\( = \left(a + \frac{1}{a}\right)^2 \)

(b) Let \( y = x + 2 \). The expression simplifies to:
\( 3y^2 + 17y + 10 \)
Splitting the middle term with product \( 3 \times 10 = 30 \) and sum \( 17 \) (numbers are 15 and 2):
\( 3y^2 + 15y + 2y + 10 = 3y(y + 5) + 2(y + 5) \)
\( = (y + 5)(3y + 2) \)
Substituting back \( y = x + 2 \):
\( = [(x + 2) + 5][3(x + 2) + 2] \)
\( = (x + 7)(3x + 6 + 2) \)
\( = (x + 7)(3x + 8) \)

(c) Grouping terms to complete a square in \( x \):
\( 4x^2 - 9y^2 + 20x + 25 = (4x^2 + 20x + 25) - 9y^2 \)
Recognizing the perfect square \( (2x + 5)^2 \):
\( = (2x + 5)^2 - (3y)^2 \)
Using the difference of squares identity \( A^2 - B^2 = (A - B)(A + B) \):
\( = (2x + 5 - 3y)(2x + 5 + 3y) \)
\( = (2x - 3y + 5)(2x + 3y + 5) \)
In simple words: Match expressions to perfect squares, substitute temporary variables to simplify quadratic forms, and group terms to use difference of squares.

Exam Tip: In part (c), look for three terms that form a perfect square trinomial before applying the difference of squares identity.