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Q8(v):
Factorise each of the following:
(v) $27p^3 – \frac{1}{216} – \frac{9}{2}p^2 + \frac{1}{4}p$
Solution :
Initial Setup & Given Expression
We are tasked with factorising the following algebraic expression:
$ 27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4}p $
To systematically approach the factorisation, we first rearrange the terms in descending order of the powers of the variable $p$ [Per standard polynomial representation conventions]:
$ 27p^3 - \frac{9}{2}p^2 + \frac{1}{4}p - \frac{1}{216} $
Step 1: Identifying the Applicable Algebraic Identity
The expression consists of four terms, beginning and ending with perfect cubes. This structural signature strongly indicates the expansion of the binomial cube identity. We recall the standard algebraic identity for the cube of a difference:
$ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $
Step 2: Extracting the Base Variables ($a$ and $b$)
We analyze the first and last terms of our rearranged polynomial to determine the values of $a$ and $b$.
- First Term ($a^3$): The term $27p^3$ can be written as a perfect cube. Since $3^3 = 27$, we have:
$ a^3 = (3p)^3 \implies a = 3p $ - Last Term ($b^3$): The term $\frac{1}{216}$ can be written as a perfect cube. Since $6^3 = 216$, we have:
$ b^3 = \left(\frac{1}{6}\right)^3 \implies b = \frac{1}{6} $
Step 3: Verifying the Middle Terms
To rigorously prove that the expression is indeed the expansion of $(3p - \frac{1}{6})^3$, we must substitute $a = 3p$ and $b = \frac{1}{6}$ into the middle terms of the identity ($-3a^2b$ and $+3ab^2$) and verify that they match the original polynomial.
Checking the second term ($-3a^2b$):
$ -3a^2b = -3(3p)^2\left(\frac{1}{6}\right) $
$ = -3(9p^2)\left(\frac{1}{6}\right) $
$ = -27p^2 \left(\frac{1}{6}\right) $
$ = -\frac{27}{6}p^2 $
Simplifying the fraction by dividing the numerator and denominator by 3:
$ = -\frac{9}{2}p^2 $
[This perfectly matches the second term of our rearranged polynomial.]
Checking the third term ($+3ab^2$):
$ +3ab^2 = 3(3p)\left(\frac{1}{6}\right)^2 $
$ = 9p\left(\frac{1}{36}\right) $
$ = \frac{9}{36}p $
Simplifying the fraction by dividing the numerator and denominator by 9:
$ = \frac{1}{4}p $
[This perfectly matches the third term of our rearranged polynomial.]
Step 4: Constructing the Factorised Form
Since all four terms of the given polynomial perfectly map to the expansion of $(a - b)^3$, we can confidently compress the expression into its factorised binomial cube form.
Substituting $a = 3p$ and $b = \frac{1}{6}$ into $(a - b)^3$ yields:
$ \left(3p - \frac{1}{6}\right)^3 $
To express this as a complete factorisation (a product of irreducible polynomials), we write the binomial three times:
$ \left(3p - \frac{1}{6}\right) \left(3p - \frac{1}{6}\right) \left(3p - \frac{1}{6}\right) $
Final Solution: The fully factorised form of the given polynomial is $ \left(3p - \frac{1}{6}\right) \left(3p - \frac{1}{6}\right) \left(3p - \frac{1}{6}\right) $.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.4
- Q1(i): Use suitable identities to find the following products: (i) $(x + 4) (x + 10)$
- Q1(ii): Use suitable identities to find the following products: (ii) $(x + 8) (x – 10)$
- Q1(iii): Use suitable identities to find the following products: (iii) $(3x + 4) (3x – 5)$
- Q1(iv): Use suitable identities to find the following products: (iv) $(y^2 + \frac{3}{2}) (y^2 – \frac{3}{2})$
- Q1(v): Use suitable identities to find the following products: (v) $(3 – 2x) (3 + 2x)$
- Q2(i): Evaluate the following products without multiplying directly: (i) $103 \times 107$
- Q2(ii): Evaluate the following products without multiplying directly: (ii) $95 \times 96$
- Q2(iii): Evaluate the following products without multiplying directly: (iii) $104 \times 96$
- Q3(i): Factorise the following using appropriate identities: (i) $9x^2 + 6xy + y^2$
- Q3(ii): Factorise the following using appropriate identities: (ii) $4y^2 – 4y + 1$
- Q3(iii): Factorise the following using appropriate identities: (iii) $x^2 – \frac{y^2}{100}$
- Q4(i): Expand each of the following, using suitable identities: (i) $(x + 2y + 4z)^2$
- Q4(ii): Expand each of the following, using suitable identities: (ii) $(2x – y + z)^2$
- Q4(iii): Expand each of the following, using suitable identities: (iii) $(–2x + 3y + 2z)^2$
- Q4(iv): Expand each of the following, using suitable identities: (iv) $(3a – 7b – c)^2$
- Q4(v): Expand each of the following, using suitable identities: (v) $(–2x + 5y – 3z)^2$
- Q4(vi): Expand each of the following, using suitable identities: (vi) $(\frac{1}{4}a - \frac{1}{2}b + 1)^2$
- Q5(i): Factorise: (i) $4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz$
- Q5(ii): Factorise: (ii) $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz$
- Q6(i): Write the following cubes in expanded form: (i) $(2x + 1)^3$
- Q6(ii): Write the following cubes in expanded form: (ii) $(2a – 3b)^3$
- Q6(iii): Write the following cubes in expanded form: (iii) $(\frac{3}{2}x + 1)^3$
- Q6(iv): Write the following cubes in expanded form: (iv) $(x - \frac{2}{3}y)^3$
- Q7(i): Evaluate the following using suitable identities: (i) $(99)^3$
- Q7(ii): Evaluate the following using suitable identities: (ii) $(102)^3$
- Q7(iii): Evaluate the following using suitable identities: (iii) $(998)^3$
- Q8(i): Factorise each of the following: (i) $8a^3 + b^3 + 12a^2b + 6ab^2$
- Q8(ii): Factorise each of the following: (ii) $8a^3 – b^3 – 12a^2b + 6ab^2$
- Q8(iii): Factorise each of the following: (iii) $27 – 125a^3 – 135a + 225a^2$
- Q8(iv): Factorise each of the following: (iv) $64a^3 – 27b^3 – 144a^2b + 108ab^2$
- Q9(i): Verify : (i) $x^3 + y^3 = (x + y) (x^2 – xy + y^2)$
- Q9(ii): Verify : (ii) $x^3 – y^3 = (x – y) (x^2 + xy + y^2)$
- Q10(i): Factorise each of the following: (i) $27y^3 + 125z^3$ [Hint : See Question 9.]
- Q10(ii): Factorise each of the following: (ii) $64m^3 – 343n^3$ [Hint : See Question 9.]
- Q11: Factorise : $27x^3 + y^3 + z^3 – 9xyz$
- Q12: Verify that $x^3 + y^3 + z^3 – 3xyz = \frac{1}{2}(x + y + z)[(x – y)^2 + (y – z)^2 + (z – x)^2]$
- Q13: If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$.
- Q14(i): Without actually calculating the cubes, find the value of each of the following: (i) $(–12)^3 + (7)^3 + (5)^3$
- Q14(ii): Without actually calculating the cubes, find the value of each of the following: (ii) $(28)^3 + (–15)^3 + (–13)^3$
- Q15(i): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (i) Area : $25a^2 – 35a + 12$
- Q15(ii): Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (ii) Area : $35y^2 + 13y –12$
- Q16(i): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume : $3x^2 – 12x$
- Q16(ii): What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (ii) Volume : $12ky^2 + 8ky – 20k$
CBSE Solutions for Class 9 Mathematics Polynomials
Chapters in CBSE - Class 9 Mathematics
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