Exercise 4.3 — Class 9 Mathematics Solutions

Solution

Solution

1. Prime and monomial factorization of each term.

   Expand each term into its prime numerical factors and irreducible variable factors:

   $$\begin{array}{cc}21x^{2}y& =3\times 7\times x\times x\times y\\ 35x{y}^2& =5\times 7\times x\times y\times y\end{array}$$

2. Identify the common prime factors.

   Comparing the factorizations of both monomials:

   • Common numerical prime factor: $7$
   • Common variable factors: $x$ and $y$

3. Compute the Highest Common Factor (HCF).

   The HCF is the product of all shared common factors:

   $$\begin{array}{cc}\text{HCF}& =7\times x\times y\\ & =7xy\end{array}$$

Solution

1. Factorize the first expression using the difference of two squares.

   Apply the algebraic identity $a^2-b^2=(a-b)(a+b)$ to $4x^2-9y^2$:

   $$\begin{array}{cc}4x^2-9y^2& =(2x{)}^2-(3y{)}^2\\ & =(2x-3y)(2x+3y)\end{array}$$

2. Factorize the second expression by taking out the common monomial factor.

   Extract the greatest common factor $x$ from both terms of $2x^2-3xy$:

   $$\begin{array}{cc}2x^2-3xy& =x(2x-3y)\end{array}$$

3. Identify the common factor to determine the HCF.

   The common binomial factor present in both expressions is $(2x-3y)$:

   $$\begin{array}{cc}\text{HCF}& =2x-3y\end{array}$$

Solution

1. Factorize the first expression using the difference of two cubes.

   Recall the algebraic identity for the difference of two cubes:

   $$\begin{array}{cc}{a}^3-b^3& =(a-b)(a^2+ab+b^2)\end{array}$$

   Applying this formula to $x^3-1=(x{)}^3-(1{)}^3$:

   $$\begin{array}{cc}{x}^3-1& =(x{)}^3-(1{)}^3\\ & =(x-1)((x{)}^2+(x)(1)+(1{)}^2)\\ & =(x-1)(x^2+x+1)\end{array}$$

2. Inspect the second expression.

   The quadratic factor $x^2+x+1$ cannot be factored further over the real numbers because its discriminant is negative ($b^2-4ac=1^2-4(1)(1)=-3<0$):

   $$\begin{array}{cc}{x}^2+x+1& =(x^2+x+1)\end{array}$$

3. Find the Highest Common Factor (HCF).

   Comparing the factors of both expressions, the common factor is $(x^2+x+1)$:

   $$\begin{array}{cc}\text{HCF}& =x^2+x+1\end{array}$$

Solution

1. Factorize the first expression.

   First, factor out the common monomial factor $a$:

   $$\begin{array}{cc}{a}^3+2a^2-3a& =a(a^2+2a-3)\end{array}$$

   Now factorize the quadratic trinomial $a^2+2a-3$ by splitting the middle term (finding two numbers whose product is $-3$ and sum is $+2$, which are $+3$ and $-1$):

   $$\begin{array}{cc}{a}^3+2a^2-3a& =a[a^2+3a-a-3]\\ & =a[a(a+3)-1(a+3)]\\ & =a(a+3)(a-1)\end{array}$$

2. Factorize the second expression.

   Factor out the common monomial factor $a$:

   $$\begin{array}{cc}2a^3+5a^2-3a& =a(2a^2+5a-3)\end{array}$$

   Factorize the quadratic trinomial $2a^2+5a-3$ by splitting the middle term (finding two numbers whose product is $2\times (-3)=-6$ and sum is $+5$, which are $+6$ and $-1$):

   $$\begin{array}{cc}2a^3+5a^2-3a& =a[2a^2+6a-a-3]\\ & =a[2a(a+3)-1(a+3)]\\ & =a(a+3)(2a-1)\end{array}$$

3. Identify common factors and determine the HCF.

   Comparing the factorized expressions:

   • First expression: $a(a+3)(a-1)$
   • Second expression: $a(a+3)(2a-1)$

   The common factors shared by both expressions are $a$ and $(a+3)$. Multiplying the common factors:

   $$\begin{array}{cc}\text{HCF}& =a(a+3)\end{array}$$

Solution

1. Factorize the first expression.

   Split the middle term of $t^2-3t-4$: two numbers with product $-4$ and sum $-3$ are $-4$ and $+1$.

   $$\begin{array}{cc}{t}^2-3t-4& =t^2-4t+t-4\\ & =t(t-4)+1(t-4)\\ & =(t-4)(t+1)\end{array}$$

2. Factorize the second expression.

   Two numbers with product $+4$ and sum $+5$ are $+4$ and $+1$.

   $$\begin{array}{cc}{t}^2+5t+4& =t^2+4t+t+4\\ & =t(t+4)+1(t+4)\\ & =(t+4)(t+1)\end{array}$$

3. Factorize the third expression.

   Use the difference of two squares $a^2-b^2=(a+b)(a-b)$:

   $$\begin{array}{cc}{t}^2-1& =(t{)}^2-(1{)}^2\\ & =(t+1)(t-1)\end{array}$$

4. Pick out the common factor.

   $$\begin{array}{cc}{t}^2-3t-4& =(t-4)(t+1)\\ t^2+5t+4& =(t+4)(t+1)\\ t^2-1& =(t-1)(t+1)\end{array}$$

   The only factor present in all three expressions is $(t+1)$.

   $$\begin{array}{cc}\text{HCF}& =t+1\end{array}$$

Solution

1. Factorize the first expression.

   Factorize the quadratic trinomial $x^2+15x+56$ by splitting the middle term (finding two numbers whose product is $56$ and sum is $15$, which are $7$ and $8$):

   $$\begin{array}{cc}{x}^2+15x+56& =x^2+7x+8x+56\\ & =x(x+7)+8(x+7)\\ & =(x+7)(x+8)\end{array}$$

2. Factorize the second expression.

   Factorize the quadratic trinomial $x^2+5x-24$ by splitting the middle term (finding two numbers whose product is $-24$ and sum is $+5$, which are $+8$ and $-3$):

   $$\begin{array}{cc}{x}^2+5x-24& =x^2+8x-3x-24\\ & =x(x+8)-3(x+8)\\ & =(x+8)(x-3)\end{array}$$

3. Factorize the third expression.

   Extract the common variable factor $x$ from $x^2+8x$:

   $$\begin{array}{cc}{x}^2+8x& =x(x+8)\end{array}$$

4. Identify the common factor to find the HCF.

   Comparing the factorized forms of all three expressions:

   • First expression: $(x+7)(x+8)$
   • Second expression: $(x+8)(x-3)$
   • Third expression: $x(x+8)$

   The common factor present in all three expressions is $(x+8)$:

   $$\begin{array}{cc}\text{HCF}& =x+8\end{array}$$

5. Check: why the answer is not $x+7$.

   $x+7$ is a factor only if the expression becomes $0$ at $x=-7$. It does so for the first expression only:

   $$\begin{array}{cc}(-7{)}^2+15(-7)+56& =0\\ (-7{)}^2+5(-7)-24& =-10\\ (-7{)}^2+8(-7)& =-7\end{array}$$

   So $x+7$ does not divide the second or the third expression and cannot be their HCF. At $x=-8$ all three expressions become $0$, which confirms $x+8$.

   [!NOTE] The answer key of the textbook prints $x+7$ for this part. For the expressions as printed, that is a misprint; the HCF is $x+8$.

Solution

Solution

1. Choose the dividend and the divisor.

   The first polynomial has degree $3$ and the second has degree $2$, so:

   $$\begin{array}{cc}\text{Dividend}& =x^3-9x^2+23x-15\\ \text{Divisor}& =x^2-4x+3\end{array}$$

2. First quotient term.

   $$\frac{x^3}{x^2}=x$$

   Multiply the divisor by $x$ and subtract:

   $$\begin{array}{cc}x(x^2-4x+3)& =x^3-4x^2+3x\\ (x^3-9x^2+23x-15)-(x^3-4x^2+3x)& =-5x^2+20x-15\end{array}$$

3. Second quotient term.

   $$\frac{-5x^2}{x^2}=-5$$

   Multiply the divisor by $-5$ and subtract:

   $$\begin{array}{cc}-5(x^2-4x+3)& =-5x^2+20x-15\\ (-5x^2+20x-15)-(-5x^2+20x-15)& =0\end{array}$$

   The quotient is $x-5$ and the remainder is $0$.

4. Conclusion.

   The divisor $x^2-4x+3$ divides the first polynomial exactly, so it is the HCF.

   $$\begin{array}{cc}\text{HCF}& =x^2-4x+3\end{array}$$

Solution

1. Take out the numerical factors.

   $$\begin{array}{cc}2x^3+2x^2+2x+2& =2(x^3+x^2+x+1)\\ 6x^3+12x^2+6x+12& =6(x^3+2x^2+x+2)\end{array}$$

   The HCF of the numbers $2$ and $6$ is $2$. Keep it aside and find the HCF of the two cubic polynomials by division.

2. First division.

   $$\begin{array}{cc}\text{Dividend}& =x^3+2x^2+x+2\\ \text{Divisor}& =x^3+x^2+x+1\end{array}$$ $$\frac{x^3}{x^3}=1$$ $$\begin{array}{cc}1\cdot (x^3+x^2+x+1)& =x^3+x^2+x+1\\ (x^3+2x^2+x+2)-(x^3+x^2+x+1)& =x^2+1\end{array}$$

   The remainder is $x^2+1$, which becomes the new divisor.

3. Second division.

   Divide $x^3+x^2+x+1$ by $x^2+1$.

   $$\frac{x^3}{x^2}=x$$ $$\begin{array}{cc}x(x^2+1)& =x^3+x\\ (x^3+x^2+x+1)-(x^3+x)& =x^2+1\end{array}$$ $$\frac{x^2}{x^2}=1$$ $$\begin{array}{cc}1\cdot (x^2+1)& =x^2+1\\ (x^2+1)-(x^2+1)& =0\end{array}$$

   The remainder is $0$, so the HCF of the two cubic polynomials is $x^2+1$.

4. Conclusion.

   Multiply by the common numerical factor $2$ kept aside in Step 1:

   $$\begin{array}{cc}\text{HCF}& =2(x^2+1)\\ & =2x^2+2\end{array}$$

Solution

1. HCF of $x^3-4x$ and $3x^2-6x$.

   $$\begin{array}{cc}3x^2-6x& =3(x^2-2x)\end{array}$$

   The number $3$ is not a factor of $x^3-4x$, so drop it and divide by $x^2-2x$.

   $$\begin{array}{cc}\text{Dividend}& =x^3-4x\\ \text{Divisor}& =x^2-2x\end{array}$$ $$\frac{x^3}{x^2}=x$$ $$\begin{array}{cc}x(x^2-2x)& =x^3-2x^2\\ (x^3-4x)-(x^3-2x^2)& =2x^2-4x\end{array}$$ $$\frac{2x^2}{x^2}=2$$ $$\begin{array}{cc}2(x^2-2x)& =2x^2-4x\\ (2x^2-4x)-(2x^2-4x)& =0\end{array}$$

   The remainder is $0$, so the HCF of these two expressions is $x^2-2x$.

2. HCF of $x^2-2x$ and $2x^3-4x^2-16x$.

   $$\begin{array}{cc}\text{Dividend}& =2x^3-4x^2-16x\\ \text{Divisor}& =x^2-2x\end{array}$$ $$\frac{2x^3}{x^2}=2x$$ $$\begin{array}{cc}2x(x^2-2x)& =2x^3-4x^2\\ (2x^3-4x^2-16x)-(2x^3-4x^2)& =-16x\end{array}$$

   The remainder is $-16x$. The number $-16$ is not a factor of $x^2-2x$, so drop it and take $x$ as the new divisor.

3. Divide $x^2-2x$ by $x$.

   $$\frac{x^2}{x}=x$$ $$\begin{array}{cc}x\cdot x& =x^2\\ (x^2-2x)-x^2& =-2x\end{array}$$ $$\frac{-2x}{x}=-2$$ $$\begin{array}{cc}-2\cdot x& =-2x\\ (-2x)-(-2x)& =0\end{array}$$

   The remainder is $0$, so the last divisor $x$ is the HCF.

4. Conclusion.

   $$\begin{array}{cc}\text{HCF}& =x\end{array}$$

Solution

Solution

1. Factorize each monomial into prime factors.

   Express the numerical coefficients in prime factors and expand the algebraic variables:

   $$\begin{array}{cc}2a^{2}b& =2\times a\times a\times b\\ 4a{b}^2& =2\times 2\times a\times b\times b\\ 6ab& =2\times 3\times a\times b\end{array}$$

2. Identify common and non-common factors.

   • Factors common to all three expressions:

   $$\begin{array}{cc}\text{Common factors}& =2\times a\times b\\ & =2ab\end{array}$$

   • Non-common factors (remaining factors in each expression):
     • From $2a^{2}b$: factor $a$
     • From $4a{b}^2$: factors $2$ and $b$ (i.e. $2b$)
     • From $6ab$: factor $3$

   $$\begin{array}{cc}\text{Product of non-common factors}& =a\times 2b\times 3\\ & =6ab\end{array}$$

3. Compute the LCM.

   Using the LCM formula:

   $$\begin{array}{cc}\text{LCM}& =(\text{Product of common factors})\times (\text{Product of non-common factors})\\ & =(2ab)\times (6ab)\\ & =12a^2{b}^2\end{array}$$

Solution

1. Factorize each polynomial by taking out common factors.

   Factor the first expression:

   $$\begin{array}{cc}{x}^2+x& =x(x+1)\end{array}$$

   Factor the second expression:

   $$\begin{array}{cc}{x}^3+x^2& =x^2(x+1)\\ & =x\times x(x+1)\end{array}$$

2. Identify common and non-common factors.

   • Common factors to both expressions:

   $$\begin{array}{cc}\text{Common factors}& =x(x+1)\end{array}$$

   • Non-common factors:

   $$\begin{array}{cc}\text{Non-common factors}& =x\end{array}$$

3. Compute the LCM.

   Using the LCM formula:

   $$\begin{array}{cc}\text{LCM}& =(\text{Product of common factors})\times (\text{Product of non-common factors})\\ & =x(x+1)\times x\\ & =x^2(x+1)\end{array}$$

Solution

1. Factorize each expression completely.

   The first expression is a perfect square trinomial of the form $x^2-2xy+y^2=(x-y{)}^2$:

   $$\begin{array}{cc}{a}^2-4a+4& =a^2-2(a)(2)+2^2\\ & =(a-2{)}^2\\ & =(a-2)(a-2)\end{array}$$

   The second expression has a common monomial factor $a$:

   $$\begin{array}{cc}{a}^2-2a& =a(a-2)\end{array}$$

2. Identify common and non-common factors.

   • Common factor to both expressions:

   $$\begin{array}{cc}\text{Common factor}& =a-2\end{array}$$

   • Non-common factors:
     • From the first expression: $(a-2)$
     • From the second expression: $a$

   $$\begin{array}{cc}\text{Product of non-common factors}& =a(a-2)\end{array}$$

3. Compute the LCM.

   Using the LCM formula:

   $$\begin{array}{cc}\text{LCM}& =(\text{Product of common factors})\times (\text{Product of non-common factors})\\ & =(a-2)\times a(a-2)\\ & =a(a-2{)}^2\end{array}$$

Solution

1. Factorize each expression completely into irreducible factors.

   For the first expression, apply the difference of two squares identity $A^2-B^2=(A+B)(A-B)$ repeatedly:

   $$\begin{array}{cc}{x}^4-16& =(x^2{)}^2-4^2\\ & =(x^2+4)(x^2-4)\\ & =(x^2+4)(x^2-2^2)\\ & =(x^2+4)(x+2)(x-2)\end{array}$$

   For the second expression, take out the common factor $x$, then apply the difference of two squares:

   $$\begin{array}{cc}{x}^3-4x& =x(x^2-4)\\ & =x(x^2-2^2)\\ & =x(x+2)(x-2)\end{array}$$

2. Identify common and non-common factors.

   • Factors common to both expressions:

   $$\begin{array}{cc}\text{Product of common factors}& =(x+2)(x-2)\\ & =x^2-4\end{array}$$

   • Non-common factors:
     • From the first expression: $(x^2+4)$
     • From the second expression: $x$

   $$\begin{array}{cc}\text{Product of non-common factors}& =x(x^2+4)\end{array}$$

3. Compute the LCM.

   Using the LCM formula:

   $$\begin{array}{cc}\text{LCM}& =(\text{Product of common factors})\times (\text{Product of non-common factors})\\ & =(x^2-4)\times x(x^2+4)\\ & =x(x^2-4)(x^2+4)\\ & =x[(x^2{)}^2-4^2]\\ & =x(x^4-16)\end{array}$$

Solution

1. Factorize each expression completely into irreducible factors.

   For the first expression, factor out 4 and express with aligned signs:

   $$\begin{array}{cc}16-4x^2& =4(4-x^2)\\ & =4(2^2-x^2)\\ & =4(2+x)(2-x)\\ & =4(x+2)[-(x-2)]\\ & =-4(x+2)(x-2)\end{array}$$

   For the second expression, factorize the quadratic trinomial by splitting the middle term:

   $$\begin{array}{cc}{x}^2+x-6& =x^2+3x-2x-6\\ & =x(x+3)-2(x+3)\\ & =(x+3)(x-2)\end{array}$$

   For the third expression, apply difference of squares and align signs:

   $$\begin{array}{cc}4-x^2& =2^2-x^2\\ & =(2+x)(2-x)\\ & =(x+2)[-(x-2)]\\ & =-1(x+2)(x-2)\end{array}$$

2. Identify common and non-common factors.

   • Factor common to all three expressions:

   $$\begin{array}{cc}\text{Common to all three}& =x-2\end{array}$$

   • Common factor between the first and third expressions:

   $$\begin{array}{cc}\text{Common between 1st and 3rd}& =-(x+2)\end{array}$$

   • Combining the common factors:

   $$\begin{array}{cc}\text{Product of common factors}& =(-1)(x+2)(x-2)\\ & =(-1)(x^2-4)\\ & =4-x^2\end{array}$$

   • Remaining non-common factors:
     • From $16-4x^2$: the numerical coefficient $4$
     • From $x^2+x-6$: $(x+3)$
     • From $4-x^2$: no remaining factor (accounted for in common factors)

   $$\begin{array}{cc}\text{Product of non-common factors}& =4\times (x+3)\\ & =4(x+3)\end{array}$$

3. Compute the LCM.

   Using the LCM formula:

   $$\begin{array}{cc}\text{LCM}& =(\text{Product of common factors})\times (\text{Product of non-common factors})\\ & =(4-x^2)\times 4(x+3)\\ & =4(4-x^2)(x+3)\end{array}$$

Solution

1. State given data and formula.

   Let the two polynomials be $p(y)$ and $q(y)$. Given:

   $$\begin{array}{cc}\text{HCF}& =y-7\\ \text{LCM}& =y^3-10y^2+11y+70\\ p(y)& =y^2-5y-14\end{array}$$

   Using the relationship between two polynomials and their HCF and LCM:

   $$\begin{array}{cc}p(y)\times q(y)& =\text{HCF}\times \text{LCM}\\ q(y)& =\frac{\text{HCF}\times \text{LCM}}{p(y)}\end{array}$$

2. Factorize the denominator polynomial $p(y)$.

   Factorize $p(y)=y^2-5y-14$ by splitting the middle term: Find two numbers whose product is $-14$ and whose sum is $-5$. The numbers are $-7$ and $2$:

   $$\begin{array}{cc}p(y)& =y^2-7y+2y-14\\ & =y(y-7)+2(y-7)\\ & =(y-7)(y+2)\end{array}$$

3. Substitute factorized values and cancel common factors.

   Substitute $\text{HCF}=y-7$, $\text{LCM}=y^3-10y^2+11y+70$, and $p(y)=(y-7)(y+2)$:

   $$\begin{array}{cc}q(y)& =\frac{\text{HCF}\times \text{LCM}}{p(y)}\\ & =\frac{(y-7)(y^3-10y^2+11y+70)}{(y-7)(y+2)}\\ & =\frac{y^3-10y^2+11y+70}{y+2}\end{array}$$

   The common binomial factor $(y-7)$ cancels from both the numerator and the denominator.

4. Perform polynomial division.

   Divide the numerator $y^3-10y^2+11y+70$ by $(y+2)$:

   1. First quotient term: Divide the leading term $y^3$ by $y$:

   $$\begin{array}{cc}\frac{y^3}{y}& =y^2\end{array}$$

   Multiply $y^2(y+2)=y^3+2y^2$ and subtract:

   $$\begin{array}{cc}(y^3-10y^2)-(y^3+2y^2)& =-12y^2\end{array}$$

   2. Second quotient term: Bring down $+11y$ to get $-12y^2+11y$. Divide $-12y^2$ by $y$:

   $$\begin{array}{cc}\frac{-12y^2}{y}& =-12y\end{array}$$

   Multiply $-12y(y+2)=-12y^2-24y$ and subtract:

   $$\begin{array}{cc}(-12y^2+11y)-(-12y^2-24y)& =35y\end{array}$$

   3. Third quotient term: Bring down $+70$ to get $35y+70$. Divide $35y$ by $y$:

   $$\begin{array}{cc}\frac{35y}{y}& =35\end{array}$$

   Multiply $35(y+2)=35y+70$ and subtract:

   $$\begin{array}{cc}(35y+70)-(35y+70)& =0\end{array}$$

   The division completes with remainder $0$. Therefore, the quotient polynomial is:

   $$\begin{array}{cc}q(y)& =y^2-12y+35\end{array}$$

5. Factorize the quotient into linear factors.

   The quotient gives the required second polynomial $q(y)$. Factorize the quadratic expression $y^2-12y+35$ by splitting the middle term:

   $$\begin{array}{cc}q(y)& =y^2-12y+35\\ & =y^2-7y-5y+35\\ & =y(y-7)-5(y-7)\\ & =(y-5)(y-7)\end{array}$$

Solution

1. State the given data and formula.

   Given:

   $$\begin{array}{cc}\text{LCM}& =36x^3(x+a)(x^3-a^3)\\ \text{HCF}& =x^2(x-a)\\ p(x)& =4x^2(x^2-a^2)\end{array}$$

   Using the fundamental relationship between two polynomials and their HCF and LCM:

   $$\begin{array}{cc}p(x)\times q(x)& =\text{HCF}\times \text{LCM}\\ q(x)& =\frac{\text{HCF}\times \text{LCM}}{p(x)}\end{array}$$

2. Substitute the given polynomials.

   $$\begin{array}{cc}q(x)& =\frac{x^2(x-a)\times 36x^3(x+a)(x^3-a^3)}{4x^2(x^2-a^2)}\\ & =\frac{36x^3\cdot x^2(x+a)(x-a)(x^3-a^3)}{4x^2(x^2-a^2)}\end{array}$$

3. Recognize algebraic identity and simplify.

   Using the difference of two squares identity $(x+a)(x-a)=x^2-a^2$:

   $$\begin{array}{cc}q(x)& =\frac{36x^3\cdot x^2(x^2-a^2)(x^3-a^3)}{4x^2(x^2-a^2)}\end{array}$$

4. Cancel common factors.

   Cancel the common factors $x^2$ and $(x^2-a^2)$ from both numerator and denominator, and divide the numerical coefficients:

   $$\begin{array}{cc}q(x)& =\left(\frac{36}{4}\right)\cdot \frac{x^2}{x^2}\cdot \frac{x^2-a^2}{x^2-a^2}\cdot x^3(x^3-a^3)\\ & =9\cdot 1\cdot 1\cdot x^3(x^3-a^3)\\ & =9x^3(x^3-a^3)\end{array}$$

   Optionally, expanding into standard polynomial form:

   $$\begin{array}{cc}q(x)& =9x^6-9a^3{x}^3\end{array}$$