CBSE Class 9 Mathematics Polynomials Assignment Set 07

POLYNOMIALS

Question 1. Factorise
(a) \(6\sqrt{3}x^2 - 47x + 5\sqrt{3}\)
(c) \(5\sqrt{5}x^2 + 20x + 3\sqrt{5}\)
Answer: For (a):
We need to split the middle term \(-47x\) into two parts whose product is equal to \((6\sqrt{3}) \times (5\sqrt{3}) = 6 \times 5 \times 3 = 90\) and whose sum is \(-47\).
The numbers are \(-45\) and \(-2\).
\(6\sqrt{3}x^2 - 45x - 2x + 5\sqrt{3}\)
\(= 3\sqrt{3}x(2x - 5\sqrt{3}) - 1(2x - 5\sqrt{3})\)
\(= (3\sqrt{3}x - 1)(2x - 5\sqrt{3})\)

For (c):
We need to split the middle term \(20x\) into two parts whose product is \((5\sqrt{5}) \times (3\sqrt{5}) = 5 \times 3 \times 5 = 75\) and whose sum is \(20\).
The numbers are \(15\) and \(5\).
\(5\sqrt{5}x^2 + 15x + 5x + 3\sqrt{5}\)
\(= 5x(\sqrt{5}x + 3) + \sqrt{5}(\sqrt{5}x + 3)\)
\(= (5x + \sqrt{5})(\sqrt{5}x + 3)\)
In simple words: Split the middle term by finding two numbers that multiply to give the product of the outer coefficients and add to give the middle coefficient.

Exam Tip: Be careful with radical arithmetic when multiplying coefficients containing square roots, such as \(\sqrt{3} \times \sqrt{3} = 3\).

Question 2. Factorise
(a) \(\frac{1}{2}x^2 + 4x + 6\)
(b) \(21x^2 - 2x + \frac{1}{21}\)
Answer: For (a):
\(\frac{1}{2}x^2 + 4x + 6 = \frac{1}{2}(x^2 + 8x + 12)\)
Find two numbers that multiply to 12 and add to 8. These are 6 and 2.
\(= \frac{1}{2}(x^2 + 6x + 2x + 12)\)
\(= \frac{1}{2}[x(x + 6) + 2(x + 6)]\)
\(= \frac{1}{2}(x + 6)(x + 2) = (x + 6)\left(\frac{x}{2} + 1\right)\)

For (b):
\(21x^2 - 2x + \frac{1}{21} = \frac{1}{21}(441x^2 - 42x + 1)\)
Notice that \(441x^2 - 42x + 1 = (21x)^2 - 2(21x)(1) + (1)^2 = (21x - 1)^2\)
\(= \frac{1}{21}(21x - 1)(21x - 1) = (21x - 1)\left(x - \frac{1}{21}\right)\)
In simple words: Take out the common denominator fraction first to make factoring the quadratic expression easier.

Exam Tip: Factoring out fractional coefficients at the beginning simplifies the numbers inside the parenthesis and avoids computational errors.

Question 3. Factorise
(a) \(16a^4 + 54a\)
(b) \(216\) a^3 - 125 \)
Answer: For (a):
\(16a^4 + 54a = 2a(8a^3 + 27)\)
Using the sum of cubes formula \(X^3 + Y^3 = (X + Y)(X^2 - XY + Y^2)\) where \(X = 2a\) and \(Y = 3\):
\(= 2a\left[(2a)^3 + (3)^3\right]\)
\(= 2a(2a + 3)\left((2a)^2 - (2a)(3) + (3)^2\right)\)
\(= 2a(2a + 3)(4a^2 - 6a + 9)\)

For (b):
\(216a^3 - 125 = (6a)^3 - (5)^3\)
Using the difference of cubes formula \(X^3 - Y^3 = (X - Y)(X^2 + XY + Y^2)\) where \(X = 6a\) and \(Y = 5\):
\(= (6a - 5)\left((6a)^2 + (6a)(5) + (5)^2\right)\)
\(= (6a - 5)(36a^2 + 30a + 25)\)
In simple words: Recognize expressions as sums or differences of cubes and apply the standard algebraic expansion formulas.

Exam Tip: Always check for a common monomial factor before applying special product formulas like difference or sum of cubes.

Question 4. Factorise
(a) \(x^3 - \frac{1}{x^3} - 2x + \frac{2}{x}\)
(b) \(27x^3 + 5\sqrt{5}y^3\)
Answer: For (a):
\(x^3 - \frac{1}{x^3} - 2x + \frac{2}{x} = \left(x^3 - \frac{1}{x^3}\right) - 2\left(x - \frac{1}{x}\right)\)
Using the expansion \(x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(x^2 + 1 + \frac{1}{x^2}\right)\):
\(= \left(x - \frac{1}{x}\right)\left(x^2 + 1 + \frac{1}{x^2}\right) - 2\left(x - \frac{1}{x}\right)\)
Factor out \(\left(x - \frac{1}{x}\right)\):
\(= \left(x - \frac{1}{x}\right)\left[x^2 + 1 + \frac{1}{x^2} - 2\right]\)
\(= \left(x - \frac{1}{x}\right)\left(x^2 - 1 + \frac{1}{x^2}\right)\)

For (b):
\(27x^3 + 5\sqrt{5}y^3 = (3x)^3 + (\sqrt{5}y)^3\)
Using the sum of cubes formula \((3x + \sqrt{5}y)((3x)^2 - (3x)(\sqrt{5}y) + (\sqrt{5}y)^2)\):
\(= (3x + \sqrt{5}y)(9x^2 - 3\sqrt{5}xy + 5y^2)\)
In simple words: Group terms intelligently to reveal common binomial factors or express terms in cube form.

Exam Tip: In part (a), grouping \(-2x + \frac{2}{x}\) as \(-2\left(x - \frac{1}{x}\right)\) is the crucial step that creates the common factor.

Question 5. Simplify
\(\frac{991 \times 991 \times 991 + 9 \times 9 \times 9}{991 \times 991 - 991 \times 9 + 9 \times 9}\)
Answer: Let \(a = 991\) and \(b = 9\).
The expression becomes:
\(\frac{a^3 + b^3}{a^2 - ab + b^2}\)
Using the algebraic identity \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\):
\(= \frac{(a + b)(a^2 - ab + b^2)}{a^2 - ab + b^2}\)
\(= a + b\)
Substitute the values of \(a\) and \(b\):
\(= 991 + 9 = 1000\)
In simple words: Use variables to substitute large numbers, simplify the algebraic fraction using identities, and then substitute back.

Exam Tip: Never perform direct multiplication with large numbers in such problems; always look for an algebraic identity.

Question 6. Factorise
\(8a^3 + 27b^3 + 64c^3 - 72abc\)
Answer: Rewrite the expression to fit the standard cube identity:
\((2a)^3 + (3b)^3 + (4c)^3 - 3(2a)(3b)(4c)\)
Using the identity \(X^3 + Y^3 + Z^3 - 3XYZ = (X + Y + Z)(X^2 + Y^2 + Z^2 - XY - YZ - ZX)\):
Where \(X = 2a\), \(Y = 3b\), and \(Z = 4c\):
\(= (2a + 3b + 4c)\left((2a)^2 + (3b)^2 + (4c)^2 - (2a)(3b) - (3b)(4c) - (4c)(2a)\right)\)
\(= (2a + 3b + 4c)(4a^2 + 9b^2 + 16c^2 - 6ab - 12bc - 8ac)\)
In simple words: Express the given polynomial in the form \(X^3 + Y^3 + Z^3 - 3XYZ\) and apply the standard expansion formula.

Exam Tip: Memorize the expanded form of the sum of three cubes identity along with its negative cross-product terms.

Question 7. Factorise
\(x^3 - 2\sqrt{2}y^3 + 3\sqrt{3}z^3 + 3\sqrt{6}xyz\)
Answer: Rewrite the expression in standard form:
\(x^3 + (-\sqrt{2}y)^3 + (\sqrt{3}z)^3 - 3(x)(-\sqrt{2}y)(\sqrt{3}z)\)
Notice that \(-3(x)(-\sqrt{2}y)(\sqrt{3}z) = 3\sqrt{6}xyz\), which matches the given expression.
Using the identity \(X^3 + Y^3 + Z^3 - 3XYZ = (X + Y + Z)(X^2 + Y^2 + Z^2 - XY - YZ - ZX)\):
Where \(X = x\), \(Y = -\sqrt{2}y\), and \(Z = \sqrt{3}z\):
\(= (x - \sqrt{2}y + \sqrt{3}z)\left((x)^2 + (-\sqrt{2}y)^2 + (\sqrt{3}z)^2 - (x)(-\sqrt{2}y) - (-\sqrt{2}y)(\sqrt{3}z) - (\sqrt{3}z)(x)\right)\)
\(= (x - \sqrt{2}y + \sqrt{3}z)(x^2 + 2y^2 + 3z^2 + \sqrt{2}xy + \sqrt{6}yz - \sqrt{3}xz)\)
In simple words: Adjust signs and radical terms to fit the sum of three cubes identity format.

Exam Tip: Pay close attention to negative signs inside cube terms like \((-\sqrt{2}y)^3 = -2\sqrt{2}y^3\).

Question 8. Factorise
\((p - q)^3 + (q - r)^3 + (r - p)^3\)
Answer: Let \(a = p - q\), \(b = q - r\), and \(c = r - p\).
Find the sum \(a + b + c\):
\(a + b + c = (p - q) + (q - r) + (r - p) = 0\)
We know the algebraic identity: if \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\).
Substitute back the values of \(a\), \(b\), and \(c\):
\((p - q)^3 + (q - r)^3 + (r - p)^3 = 3(p - q)(q - r)(r - p)\)
In simple words: When the sum of three bases is zero, the sum of their cubes equals three times their product.

Exam Tip: Always check if the sum of the terms inside the cubes is zero before applying the special conditional identity.

Question 9. Without actual calculating, find
\((28)^3 + (-15)^3 + (-13)^3\)
Answer: Let \(a = 28\), \(b = -15\), and \(c = -13\).
Find the sum \(a + b + c\):
\(a + b + c = 28 + (-15) + (-13) = 28 - 28 = 0\)
Since \(a + b + c = 0\), we use the identity \(a^3 + b^3 + c^3 = 3abc\):
\((28)^3 + (-15)^3 + (-13)^3 = 3(28)(-15)(-13)\)
Calculate the product step-by-step:
\(3 \times 28 = 84\)
\(84 \times (-15) = -1260\)
\(-1260 \times (-13) = 16380\)
In simple words: Verify that the sum of the numbers is zero, then use the property \(a^3 + b^3 + c^3 = 3abc\).

Exam Tip: State the conditional identity clearly before substituting the numerical values.

Question 10. Evaluate by using suitable identity
(a) \((102)^3\)
(b) \((998)^3\)
Answer: For (a):
\((102)^3 = (100 + 2)^3\)
Using the identity \((X + Y)^3 = X^3 + Y^3 + 3XY(X + Y)\):
\(= (100)^3 + (2)^3 + 3(100)(2)(100 + 2)\)
\(= 1000000 + 8 + 600(102)\)
\(= 1000008 + 61200 = 1061208\)

For (b):
\((998)^3 = (1000 - 2)^3\)
Using the identity \((X - Y)^3 = X^3 - Y^3 - 3XY(X - Y)\):
\(= (1000)^3 - (2)^3 - 3(1000)(2)(1000 - 2)\)
\(= 1000000000 - 8 - 6000(998)\)
\(= 999999992 - 5988000 = 994011992\)
In simple words: Split numbers into base tens (like \(100 + 2\) or \(1000 - 2\)) to easily apply cube binomial formulas.

Exam Tip: Choose numbers close to powers of 10 to make arithmetic multiplication simple and error-free.