Maharashtra Board Class 9 Maths Chapter 3 Set 3.6 Algebra Standard Part 1 Polynomials Solutions

Question 1. Find the factors of the polynomials given below:
(i) \(2x^2 + x - 1\)
(ii) \(2m^2 + 5m - 3\)
(iii) \(12x^2 + 61x + 77\)
(iv) \(3y^2 - 2y - 1\)
(v) \(\sqrt{3}x^2 + 4x + \sqrt{3}\)
(vi) \(\frac{1}{2}x^2 - 3x + 4\)
Answer:
(i) Solution:
\(2x^2 + x - 1\)
\( = 2x^2 + 2x - x - 1 \)
\( = 2x(x + 1) - 1(x + 1) \)
\( = (x + 1)(2x - 1) \)
\( 2 \times -1 = -2 \)
\( 2 \)
\( -1 \)
\( 2 \times -1 = -2 \)
\( 2 - 1 = 1 \)
(ii) Solution:
\(2m^2 + 5m - 3\)
\( = 2m^2 + 6m - m - 3 \)
\( = 2m(m + 3) - 1(m + 3) \)
\( = (m + 3)(2m - 1) \)
\( 2 \times -3 = -6 \)
\( 6 \)
\( -1 \)
\( 6 \times -1 = -6 \)
\( 6 - 1 = 5 \)
(iii) Solution:
\(12x^2 + 61x + 77\)
\( = 12x^2 + 28x + 33x + 77 \)
\( = 4x(3x + 7) 4 + 11(3x + 7) \)
\( = (3x + 7)(4x + 11) \)
\( 12 \times 77 = 4 \times 3 \times 11 \times 7 \)
\( = 28 \times 33 \)
\( 28 \qquad 33 \)
\( 28 + 33 = 61 \)
\( 28 \times 33 = 924 \)
(iv) Solution:
\(3y^2 - 2y - 1\)
\( = 3y^2 - 3y + y - 1 \)
\( = 3y(y - 1) + 1(y - 1) \)
\( = (y - 1)(3y + 1) \)
\( 3 \times -1 = -3 \)
\( -3 \qquad 1 \)
\( -3 \times 1 = -3 \)
\( -3 + 1 = -2 \)
(v) Solution:
\(\sqrt{3}x^2 + 4x + \sqrt{3}\)
\( = \sqrt{3}x^2 + 3x + x + \sqrt{3} \)
\( = \sqrt{3}x^2 + \sqrt{3} \times \sqrt{3}x + x + \sqrt{3} \)
\( = \sqrt{3}x(x + \sqrt{3}) + 1(x + \sqrt{3}) \)
\( = (x + \sqrt{3})(\sqrt{3}x + 1) \)
\( \sqrt{3} \times \sqrt{3} = 3 \)
\( 3 \qquad 1 \)
\( 3 \times 1 = 3 \)
\( 3 + 1 = 4 \)
(vi) Solution:
\(\frac{1}{2}x^2 - 3x + 4\)
\( = \frac{1}{2}x^2 - 2x - x + 4 \)
\( = \frac{1}{2}x^2 - \frac{2}{2}x - x + 4 \)
\( = \frac{1}{2}x(x - 4) - 1(x - 4) \)
\( = (x - 4)(\frac{1}{2}x - 1) \)
\( \frac{1}{2} \times 4 = 2 \)
\( -2 \qquad -1 \)
\( -2 \times -1 = 2 \)
\( -2 - 1 = -3 \)
Alternate method
\(\frac{1}{2}x^2 - 3x + 4 = \frac{1}{2}(x^2 - 6x + 8)\)
\( = \frac{1}{2}(x^2 - 4x - 2x + 8)\)
\( = \frac{1}{2}[x(x - 4) - 2(x - 4)]\)
\( = \frac{1}{2}(x - 2)(x - 4)\)
\( 8 \)
\( -4 \qquad -2 \)
\( -4 \times -2 = 8 \)
\( -4 - 2 = -6 \)
In simple words: Factoring polynomials involves rewriting the polynomial as a product of simpler expressions, typically binomials. This is often done by splitting the middle term, finding common factors, and grouping terms to reveal the binomial factors.

🎯 Exam Tip: Pay close attention to the signs and coefficients when splitting the middle term. Ensure that both the product and sum conditions are met by the chosen factors for accurate factorization.

Question 2. Factorize the following polynomials.
(i) \((x^2 - x)^2 - 8(x^2 - x) + 12\)
(iii) \((x^2 - 6x)^2 - 8(x^2 - 6x + 8) - 64\)
(v) \((y + 2)(y - 3)(y + 8)(y + 3) + 56\)
(vii) \((x - 3)(x - 4)^2(x - 5) - 6\)
Answer:
(i) Solution:
\((x^2 - x)^2 - 8(x^2 - x) + 12\)
\( = m^2 - 8m + 12 \)...[Putting \(x^2 - x = m\)]
\( = m^2 - 6m - 2m + 12 \)
\( = m(m - 6) - 2(m - 6) \)
\( = (m - 6)(m - 2) \)
\( = (x^2 - x - 6)(x^2 - x - 2) \)...[Replacing \(m = x^2 - x\)]
\( = (x^2 - 3x + 2x - 6)(x^2 - 2x + x - 2) \)
\( = [x(x - 3) + 2(x - 3)][x(x - 2) + 1(x - 2)] \)
\( = (x - 3)(x + 2)(x - 2)(x + 1) \)
(ii) Solution:
\((x - 5)^2 - (5x - 25) - 24\)
\( = (x - 5)^2 - (5x - 25) - 24 \)
\( = (x - 5)^2 - 5(x - 5) - 24 \)
\( = m^2 - 5m - 24 \)...[Putting \(x - 5 = m\)]
\( = m^2 - 8m + 3m - 24 \)
\( = m(m - 8) + 3(m - 8) \)
\( = (m - 8)(m + 3) \)
\( = (x - 5 - 8)(x - 5 + 3) \)...[Replacing \(m = x - 5\)]
\( = (x - 13)(x - 2) \)
\( -24 \)
\( -8 \qquad 3 \)
\( -8 + 3 = -5 \)
\( -8 \times 3 = -24 \)
(iii) Solution:
\((x^2 - 6x)^2 - 8(x^2 - 6x + 8) - 64\)
\( = m^2 - 8(m + 8) - 64 \)...[Putting \(x^2 - 6x = m\)]
\( = m^2 - 8m - 64 - 64 \)
\( = m^2 - 8m - 128 \)
\( = m^2 - 16m + 8m - 128 \)
\( = m(m - 16) + 8(m - 16) \)
\( = (m - 16)(m + 8) \)
\( = (x^2 - 6x - 16)(x^2 - 6x + 8) \)...[Replacing \(m = x^2 - 6x\)]
\( = (x^2 - 8x + 2x - 16)(x^2 - 4x - 2x + 8) \)
\( = [x(x - 8) + 2(x - 8)][x(x - 4) - 2(x - 4)] \)
\( = (x - 8)(x + 2)(x - 4)(x - 2) \)
(iv) Solution:
\((x^2 - 2x + 3)(x^2 - 2x + 5) - 35\)
\( = (m + 3)(m + 5) - 35 \)...[Putting \(x^2 - 2x = m\)]
\( = m(m + 5) + 3(m + 5) - 35 \)
\( = m^2 + 5m + 3m + 15 - 35 \)
\( = m^2 + 8m - 20 \)
\( = m^2 + 10m - 2m - 20 \)
\( = m(m + 10) - 2(m + 10) \)
\( = (m + 10)(m - 2) \)
\( = (x^2 - 2x + 10)(x^2 - 2x - 2) \)...[Replacing \(m = x^2 - 2x\)]
\( -20 \)
\( 10 \qquad -2 \)
\( 10 \times -2 = -20 \)
\( 10 - 2 = 8 \)
(v) Solution:
\((y + 2)(y - 3)(y + 8)(y + 3) + 56\)
\( = (y + 2)(y + 3)(y - 3)(y + 8) + 56 \)
\( = (y^2 + 3y + 2y + 6)(y^2 + 8y - 3y - 24) + 56 \)
\( = (y^2 + 5y + 6)(y^2 + 5y - 24) + 56 \)
\( = (m + 6)(m - 24) + 56 \)...[Putting \(y^2 + 5y = m\)]
\( = m(m - 24) + 6(m - 24) + 56 \)
\( = m^2 - 24m + 6m - 144 + 56 \)
\( = m^2 - 18m - 88 \)
\( = m^2 - 22m + 4m - 88 \)
\( = m(m - 22) + 4(m - 22) \)
\( = (m - 22)(m + 4) \)
\( = (y^2 + 5y - 22)(y^2 + 5y + 4) \)...[Replacing \(m = y^2 + 5y\)]
\( = (y^2 + 5y - 22)(y^2 + 4y + y + 4) \)
\( = (y^2 + 5y - 22)[y(y + 4) + 1(y + 4)] \)
\( = (y^2 + 5y - 22)(y + 4)(y + 1) \)
(vi) Solution:
\((y^2 + 5y)(y^2 + 5y - 2) - 24\)
\( = (m)(m - 2) - 24 \)...[Putting \(y^2 + 5y = m\)]
\( = m^2 - 2m - 24 \)
\( = m^2 - 6m + 4m - 24 \)
\( = m(m - 6) + 4(m - 6) \)
\( = (m - 6)(m + 4) \)
\( = (y^2 + 5y - 6)(y^2 + 5y + 4) \)...[Replacing \(m = y^2 + 5y\)]
\( = (y^2 + 6y - y - 6)(y^2 + 4y + y + 4) \)
\( = [y(y + 6) - 1(y + 6)][y(y + 4) + 1(y + 4)] \)
\( = (y + 6)(y - 1)(y + 4)(y + 1) \)
(vii) Solution:
\((x - 3)(x - 4)^2(x - 5) - 6\)
\( = (x - 3)(x - 5)(x - 4)^2 - 6 \)
\( = (x^2 - 5x - 3x + 15)(x^2 - 8x + 16) - 6 \)
\( = (x^2 - 8x + 15)(x^2 - 8x + 16) - 6 \)
\( = (m + 15)(m + 16) - 6 \)...[Putting \(x^2 - 8x = m\)]
\( = m(m + 16) + 15(m + 16) - 6 \)
\( = m^2 + 16m + 15m + 240 - 6 \)
\( = m^2 + 31m + 234 \)
\( = m^2 + 18m + 13m + 234 \)
\( = m(m + 18) + 13(m + 18) \)
\( = (m + 18)(m + 13) \)
\( = (x^2 - 8x + 18)(x^2 - 8x + 13) \)...[Replacing \(m = x^2 - 8x\)]
In simple words: Factorizing polynomials involving repeated expressions often benefits from substitution, where a complex term is replaced by a single variable (like 'm') to simplify the expression, factor it, and then substitute the original term back in. This reduces the complexity of the polynomial and makes factoring easier.

🎯 Exam Tip: When using substitution, always remember to substitute the original expression back in at the end to get the final factors in terms of the original variable. Grouping terms strategically is also key for complex polynomials.