CBSE Class 9 Mathematics Polynomials Assignment Set 08

POLYNOMIALS

Q1. Factorise \(6x^2 + 17x + 5\) by splitting the middle term.
Answer: Given expression: \(6x^2 + 17x + 5\)
Find two numbers whose product is \(6 \times 5 = 30\) and whose sum is \(17\). These numbers are \(15\) and \(2\).
Rewrite the middle term using \(15x\) and \(2x\):
\(6x^2 + 15x + 2x + 5\)
Group terms to factor out common factors:
\(3x(2x + 5) + 1(2x + 5)\)
Factor out \((2x + 5)\):
\(= (3x + 1)(2x + 5)\)
In simple words: Split the middle term into two parts that multiply to give the product of the outer terms and add up to the middle coefficient.

Exam Tip: Always check if the product of the coefficient of \(x^2\) and the constant term matches the product of your split terms.

Q2. Factorise
(a) \(4x^2 + 20xy + 25y^2\)
(b) \(7x^4 - 98x^2 + 343\)
(c) \(-1 - 2xy + x^2 + y^2\)
(d) \(12xy - 4x^2 + 1 - 9y^2\)
(e) \((a + b)^3 - a - b\)
Answer:
(a) \(4x^2 + 20xy + 25y^2\)
\(= (2x)^2 + 2(2x)(5y) + (5y)^2\)
Using the identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (2x + 5y)^2\)

(b) \(7x^4 - 98x^2 + 343\)
Factor out \(7\):
\(= 7(x^4 - 14x^2 + 49)\)
\(= 7((x^2)^2 - 2(x^2)(7) + 7^2)\)
Using the identity \((a - b)^2 = a^2 - 2ab + b^2\):
\(= 7(x^2 - 7)^2\)

(c) \(-1 - 2xy + x^2 + y^2\)
Rearrange terms:
\(= x^2 - 2xy + y^2 - 1\)
\(= (x - y)^2 - 1^2\)
Using the identity \(a^2 - b^2 = (a - b)(a + b)\):
\(= (x - y - 1)(x - y + 1)\)

(d) \(12xy - 4x^2 + 1 - 9y^2\)
Rearrange terms:
\(= 1 - (4x^2 - 12xy + 9y^2)\)
\(= 1^2 - ((2x)^2 - 2(2x)(3y) + (3y)^2)\)
\(= 1^2 - (2x - 3y)^2\)
Using the identity \(a^2 - b^2 = (a - b)(a + b)\):
\(= (1 - (2x - 3y))(1 + (2x - 3y))\)
\(= (1 - 2x + 3y)(1 + 2x - 3y)\)

(e) \((a + b)^3 - a - b\)
Factor out \(-1\) from the last two terms:
\(= (a + b)^3 - (a + b)\)
Factor out \((a + b)\):
\(= (a + b)[(a + b)^2 - 1]\)
Using the identity \(x^2 - y^2 = (x - y)(x + y)\):
\(= (a + b)(a + b - 1)(a + b + 1)\)
In simple words: Apply standard algebraic identities like square of a sum, square of a difference, and difference of squares.

Exam Tip: Rearranging terms to form recognizable groupings or standard algebraic identities is key to solving multi-variable factorisation problems.

Q3. Factorise \(x^2 - x - 6\) by using factor theorem
Answer: Let \(p(x) = x^2 - x - 6\).
The factors of the constant term \(-6\) are \(\pm 1, \pm 2, \pm 3, \pm 6\).
Test \(x = 3\):
\(p(3) = (3)^2 - (3) - 6 = 9 - 3 - 6 = 0\)
By the Factor Theorem, since \(p(3) = 0\), \((x - 3)\) is a factor of \(p(x)\).
Test \(x = -2\):
\(p(-2) = (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0\)
Since \(p(-2) = 0\), \((x + 2)\) is a factor of \(p(x)\).
Therefore, the factorised form is:
\(= (x - 3)(x + 2)\)
In simple words: Find a value that makes the polynomial zero, use the factor theorem to get the first factor, and then find the second factor.

Exam Tip: Always test factors of the constant term when applying the factor theorem to cubic or quadratic polynomials.

Q4. Factorise \(x^3 - 23x^2 + 142x - 120\)
Answer: Let \(p(x) = x^3 - 23x^2 + 142x - 120\).
Factors of constant term \(-120\) include \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 10, \pm 12\), etc.
Test \(x = 1\):
\(p(1) = (1)^3 - 23(1)^2 + 142(1) - 120 = 1 - 23 + 142 - 120 = 143 - 143 = 0\)
Since \(p(1) = 0\), \((x - 1)\) is a factor.
Perform polynomial division or synthetic division of \(p(x)\) by \((x - 1)\) to find the quotient quadratic polynomial:
\(x^3 - 23x^2 + 142x - 120 = (x - 1)(x^2 - 22x + 120)\)
Now, factorise the quadratic polynomial \(x^2 - 22x + 120\) by splitting the middle term. Find two numbers whose product is \(120\) and sum is \(-22\). These numbers are \(-10\) and \(-12\):
\(x^2 - 10x - 12x + 120 = x(x - 10) - 12(x - 10) = (x - 10)(x - 12)\)
Thus, the complete factorisation is:
\(= (x - 1)(x - 10)(x - 12)\)
In simple words: Find one linear factor using trial and error, divide the polynomial to get a quadratic expression, and then factorise that quadratic.

Exam Tip: For cubic polynomials, once you find the first linear factor, long division helps quickly reduce it to a quadratic equation.

Q5. Factorise \(x^3 - 10x^2 - 53x - 42\)
Answer: Let \(p(x) = x^3 - 10x^2 - 53x - 42\).
Factors of \(-42\) include \(\pm 1, \pm 2, \pm 3, \pm 6, \pm 7\), etc.
Test \(x = -1\):
\(p(-1) = (-1)^3 - 10(-1)^2 - 53(-1) - 42 = -1 - 10 + 53 - 42 = -53 + 53 = 0\)
Since \(p(-1) = 0\), \((x + 1)\) is a factor.
Divide \(p(x)\) by \((x + 1)\):
\(x^3 - 10x^2 - 53x - 42 = (x + 1)(x^2 - 11x - 42)\)
Now, factorise the quadratic polynomial \(x^2 - 11x - 42\) by splitting the middle term. Find two numbers whose product is \(-42\) and sum is \(-11\). These numbers are \(-14\) and \(3\):
\(x^2 - 14x + 3x - 42 = x(x - 14) + 3(x - 14) = (x - 14)(x + 3)\)
Thus, the complete factorisation is:
\(= (x + 1)(x + 3)(x - 14)\)
In simple words: Guess a root to get the first factor, divide the cubic polynomial, and factorise the remaining quadratic part.

Exam Tip: Be careful with signs when substituting negative integers into polynomials during root testing.

Q6. Factorise \(x^3 - 6x^2 + 3x + 10\) given that \((x + 1)\) is a factor
Answer: Let \(p(x) = x^3 - 6x^2 + 3x + 10\).
Since \((x + 1)\) is given as a factor, divide \(p(x)\) by \((x + 1)\):
\(x^3 - 6x^2 + 3x + 10 = (x + 1)(x^2 - 7x + 10)\)
Now, factorise the quadratic polynomial \(x^2 - 7x + 10\) by splitting the middle term. Find two numbers whose product is \(10\) and sum is \(-7\). These numbers are \(-5\) and \(-2\):
\(x^2 - 5x - 2x + 10 = x(x - 5) - 2(x - 5) = (x - 5)(x - 2)\)
Thus, the complete factorisation is:
\(= (x + 1)(x - 2)(x - 5)\)
In simple words: Since one factor is already given, use polynomial division directly to find the quadratic factor, then factorise it.

Exam Tip: When a factor is provided, you save time by skipping the trial-and-error step and proceeding directly to division.

Q7. Factorise
(a) \((x + 1)^2 - (y - 1)^2\)
(b) \(16x^4 - 81y^4\)
Answer:
(a) \((x + 1)^2 - (y - 1)^2\)
Using the identity \(a^2 - b^2 = (a - b)(a + b)\) where \(a = (x + 1)\) and \(b = (y - 1)\):
\(= [(x + 1) - (y - 1)][(x + 1) + (y - 1)]\)
\(= [x + 1 - y + 1][x + 1 + y - 1]\)
\(= (x - y + 2)(x + y)\)

(b) \(16x^4 - 81y^4\)
\(= (4x^2)^2 - (9y^2)^2\)
Using the identity \(a^2 - b^2 = (a - b)(a + b)\):
\(= (4x^2 - 9y^2)(4x^2 + 9y^2)\)
Further factorise the first term \((4x^2 - 9y^2)\) as \((2x)^2 - (3y)^2\):
\(= ((2x)^2 - (3y)^2)(4x^2 + 9y^2)\)
\(= (2x - 3y)(2x + 3y)(4x^2 + 9y^2)\)
In simple words: Apply the difference of squares formula repeatedly wherever possible.

Exam Tip: Always check if terms inside squared expressions can be further broken down using \(a^2 - b^2\).

Q8. Factorise: \((a + b)^2 - 14c(a + b) + 49c^2\)
Answer: Let \(x = (a + b)\). The expression becomes:
\(= x^2 - 14cx + 49c^2\)
\(= x^2 - 2(x)(7c) + (7c)^2\)
Using the identity \((A - B)^2 = A^2 - 2AB + B^2\):
\(= (x - 7c)^2\)
Substitute back \(x = (a + b)\):
\(= (a + b - 7c)^2\)
In simple words: Substitute a temporary variable to simplify the expression into a standard square formula, then substitute back.

Exam Tip: Substitution method simplifies complex algebraic expressions and prevents calculation errors.

Q9. Factorise: \(\left(x^2 + \frac{1}{x^2}\right) - 4\left(x + \frac{1}{x}\right) + 6\)
Answer: We know that \(\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2\), which means:
\(x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2\)
Substitute this into the given expression:
\(= \left[\left(x + \frac{1}{x}\right)^2 - 2\right] - 4\left(x + \frac{1}{x}\right) + 6\)
\(= \left(x + \frac{1}{x}\right)^2 - 4\left(x + \frac{1}{x}\right) + 4\)
Let \(y = x + \frac{1}{x}\). The expression becomes:
\(= y^2 - 4y + 4\)
\(= (y - 2)^2\)
Substitute back \(y = x + \frac{1}{x}\):
\(= \left(x + \frac{1}{x} - 2\right)^2\)
In simple words: Express the reciprocal sum squared in terms of the single reciprocal sum, then use substitution to factorise.

Exam Tip: Memorize the relation between \(x^2 + \frac{1}{x^2}\) and \(\left(x + \frac{1}{x}\right)^2\) for reciprocal polynomial problems.

Q10. Factorise: \(2x^2 + 2\sqrt{6}xy + 3y^2\)
Answer: Rewrite the expression using square roots:
\(= (\sqrt{2}x)^2 + 2(\sqrt{2}x)(\sqrt{3}y) + (\sqrt{3}y)^2\)
Using the identity \((A + B)^2 = A^2 + 2AB + B^2\) where \(A = \sqrt{2}x\) and \(B = \sqrt{3}y\):
\(= (\sqrt{2}x + \sqrt{3}y)^2\)
In simple words: Express coefficients as squares of square roots to fit the standard algebraic identity.

Exam Tip: Do not panic when irrational numbers like square roots appear in algebraic identities; treat them as standard constants.