Review Exercise 4 — Class 9 Mathematics Solutions

Solution

Solution

1. Identify the common factor.

   Find the greatest common factor $(\text{GCF})$ of the terms $12x$ and $36$:

   $$\begin{array}{cc}12x& =12\times x\\ 36& =12\times 3\end{array}$$

   The common factor is $12$.

2. Factor out $12$.

   $$\begin{array}{cc}12x+36& =12(x)+12(3)\\ & =12(x+3)\end{array}$$

3. Analyze distractors.

   • Option (b): $12(3x)=36x\ne 12x+36$
   • Option (c): $12(3x+1)=36x+12\ne 12x+36$
   • Option (d): $x(12+36x)=12x+36x^2\ne 12x+36$

   Thus, the correct choice is option (a).

Solution

1. Read the expression correctly.

   The book prints the middle term as $-12y$, but every option contains only $x$, and no product of these options can give a $y$-term. The middle term is meant to be $-12x$, so we factorize $4x^2-12x+9$.

2. Recognize the algebraic identity.

   Recall the perfect square expansion formula:

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

3. Rewrite the terms in square form.

   $$\begin{array}{cc}4x^2-12x+9& =(2x{)}^2-2(2x)(3)+3^2\\ & =(2x-3{)}^2\end{array}$$

4. Analyze distractors.

   • Option (a): $(2x+3{)}^2=4x^2+12x+9$ (middle term has $+12x$ instead of $-12x$)
   • Option (c): $(2x-3)(2x+3)=4x^2-9$ (difference of two squares)
   • Option (d): a cubic expression, so it cannot equal a quadratic

   Thus, the correct choice is option (b).

Solution

1. Express each monomial in prime power factors.

   The given expressions are:

   $$\begin{array}{cc}\text{First term}& =a^3{b}^3\\ \text{Second term}& =a^1{b}^2\end{array}$$

2. Take the lowest exponent of each common variable.

   To find the Highest Common Factor $(\text{HCF})$:

   • For base $a$: $\min (3,1)=1⟹a^1=a$
   • For base $b$: $\min (3,2)=2⟹b^2$

   $$\begin{array}{cc}\text{HCF}& =a^1{b}^2\\ & =a{b}^2\end{array}$$

3. Analyze distractors.

   • Option (a): $a^3{b}^3$ is the LCM of the variables.
   • Option (c): $a^4{b}^5$ is the product $a^3{b}^3\times a{b}^2$.
   • Option (d): $a^{2}b$ does not divide either term completely.

   Thus, the correct choice is option (b).

Solution

1. Find the LCM of the numerical coefficients.

   Prime factorization of the numbers $16$, $4$, and $30$:

   $$\begin{array}{cc}16& =2^4\\ 4& =2^2\\ 30& =2\times 3\times 5\end{array}$$

   Taking the highest power of each prime factor:

   $$\begin{array}{cc}\text{LCM}(16,4,30)& =2^4\times 3^1\times 5^1\\ & =16\times 15\\ & =240\end{array}$$

2. Find the LCM of the variable parts.

   • For base $x$: highest power among $x^2,x^1,x^1$ is $x^2$.
   • For base $y$: highest power among $y^0,y^0,y^1$ is $y$.

   Combining numerical and algebraic factors:

   $$\begin{array}{cc}\text{LCM}& =240\times x^2\times y\\ & =240x^{2}y\end{array}$$

3. Analyze distractors.

   • Option (a): $480x^{3}y$ unnecessarily doubles the coefficient and cubes $x$.
   • Option (b): $240xy$ misses the required power $x^2$.
   • Option (d): $120x^{4}y$ has an incorrect numerical multiple and power of $x$.

   Thus, the correct choice is option (c).

Solution

1. State the fundamental relationship.

   For any two algebraic polynomials $P(x)$ and $Q(x)$, their relationship with their Least Common Multiple (LCM) and Highest Common Factor (HCF) is given by:

   $$\begin{array}{cc}\text{LCM}\times \text{HCF}& =P(x)\times Q(x)\end{array}$$

2. Interpret the identity in words.

   The left-hand side is the product of LCM and HCF. The right-hand side is the product of the two polynomials. Therefore:

   $$\begin{array}{cc}\text{Product of LCM and HCF}& =\text{Product of the two polynomials}\end{array}$$

   Thus, the correct choice is option (c).

Solution

1. Factorize the quadratic expression as a perfect square.

   Apply the identity $a^2-2ab+b^2=(a-b{)}^2$:

   $$\begin{array}{cc}{x}^2-6x+9& =(x{)}^2-2(x)(3)+3^2\\ & =(x-3{)}^2\end{array}$$

2. Take the square root.

   In algebra, taking the square root of an expression yields both positive and negative branches:

   $$\begin{array}{cc}\sqrt{x^2-6x+9}& =\sqrt{(x-3{)}^2}\\ & =\pm (x-3)\end{array}$$

3. Analyze distractors.

   • Option (b): Has $+3$ inside the bracket, which corresponds to $x^2+6x+9$.
   • Options (c) and (d): Omit the essential $\pm$ sign required for complete algebraic square roots.

   Thus, the correct choice is option (a).

Solution

1. Understand LCM of identical expressions with different powers.

   When two polynomial expressions share the same common base $(a-b)$:

   • Their HCF is the base raised to the smallest exponent.
   • Their LCM is the base raised to the greatest exponent.

2. Compare the exponents.

   The exponents are $2$ and $4$:

   $$\begin{array}{cc}\text{LCM}& =(a-b{)}^{\max (2,4)}\\ & =(a-b{)}^4\end{array}$$

3. Analyze distractors.

   • Option (a): $(a-b{)}^2$ is the HCF, not the LCM.
   • Option (b): $(a-b{)}^3$ is not an exact common multiple.
   • Option (d): $(a-b{)}^6$ is the product $(a-b{)}^2\times (a-b{)}^4$, not the least common multiple.

   Thus, the correct choice is option (c).

Solution

1. Recall the cubic binomial expansion identity.

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

2. Match the given polynomial with the identity.

   Let $a=x$ and $b=1$:

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

3. Analyze distractors.

   • Option (b): $(x-1{)}^3=x^3-3x^2+3x-1$ (alternating signs)
   • Option (c): $(x+1)(x^2+x+1)=x^3+2x^2+2x+1\ne x^3+3x^2+3x+1$
   • Option (d): Expands to $x^3-2x^2+2x-1$

   Thus, the correct choice is option (a).

Solution

1. Define polynomial classifications by degree.

   In algebra, polynomials in a single variable are classified by their highest exponent (degree):

   • Degree $1$: Linear polynomial (e.g., $ax+b$)
   • Degree $2$: Quadratic polynomial (e.g., $a{x}^2+bx+c$)
   • Degree $3$: Cubic polynomial (e.g., $a{x}^3+b{x}^2+cx+d$)
   • Degree $4$: Biquadratic or quartic polynomial (e.g., $a{x}^4+\dots$)

2. Conclude the degree of a cubic polynomial.

   $$\begin{array}{cc}\text{Degree of cubic polynomial}& =3\end{array}$$

   Thus, the correct choice is option (c).

Solution

1. Recall the difference of cubes algebraic identity.

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

2. Factorize $x^3-27$.

   Write $27$ as $3^3$:

   $$\begin{array}{cc}{x}^3-27& =(x{)}^3-(3{)}^3\\ & =(x-3)((x{)}^2+(x)(3)+3^2)\\ & =(x-3)(x^2+3x+9)\end{array}$$

   The two true factors of $x^3-27$ are $(x-3)$ and $(x^2+3x+9)$.

3. Scrutinize the given options carefully.

   • Option (a): $x-3$ is a correct factor.
   • Option (b): $x+3$ is not a factor.
   • Option (c): $x^2-3x+9$ has a negative middle term $-3x$, whereas the true factor has $+3x$. Therefore, $x^2-3x+9$ is not a factor!
   • Option (d): "Both a and c" is a distractor trap based on misreading the sign in option (c).

   Because only $(x-3)$ is a valid factor among the options, option (a) is the only correct answer.

Solution

Solution

1. Identify the greatest common factor (GCF).

   Examine the numerical coefficients and variable powers of each term in the polynomial:

   • The numerical coefficients are $4$, $18$, and $-12$. Their greatest common divisor is $2$.
   • The variable factors are $x^3$, $x^2$, and $x$. The lowest power common to all terms is $x$.

   Hence, the greatest common monomial factor is $2x$:

   $$\begin{array}{cc}4x^3& =2x\times 2x^2\\ 18x^2& =2x\times 9x\\ -12x& =2x\times (-6)\end{array}$$

2. Factor out the common monomial factor.

   Using the distributive property, factor out $2x$ from the polynomial:

   $$\begin{array}{cc}4x^3+18x^2-12x& =2x(2x^2)+2x(9x)-2x(6)\\ & =2x(2x^2+9x-6)\end{array}$$

3. Check whether the quadratic factor can be factored further.

   Examine the quadratic factor $2x^2+9x-6$ over the integers: Comparing with $a{x}^2+bx+c$, we have $a=2$, $b=9$, and $c=-6$. Calculate the discriminant $D$:

   $$\begin{array}{cc}D& =b^2-4ac\\ & =(9{)}^2-4(2)(-6)\\ & =81+48\\ & =129\end{array}$$

   Since $129$ is not a perfect square, $2x^2+9x-6$ has no rational or integer roots and cannot be factored further over the integers.

   Thus, the complete factorization over integers is:

   $$\begin{array}{cc}4x^3+18x^2-12x& =2x(2x^2+9x-6)\end{array}$$

Solution

1. Express each term as a perfect cube.

   Notice that both terms are perfect cubes:

   $$\begin{array}{cc}{x}^3& =(x{)}^3\\ 64y^3& =(4y{)}^3\end{array}$$

   Rewrite the given expression as a sum of two cubes:

   $$\begin{array}{cc}{x}^3+64y^3& =(x{)}^3+(4y{)}^3\end{array}$$

2. Apply the sum of cubes identity.

   Recall the algebraic identity for the sum of two cubes:

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

   Setting $a=x$ and $b=4y$:

   $$\begin{array}{cc}{x}^3+64y^3& =(x+4y)[(x{)}^2-(x)(4y)+(4y{)}^2]\\ & =(x+4y)(x^2-4xy+16y^2)\end{array}$$

Solution

1. Express each term as a perfect cube.

   Notice that both terms are perfect cubes:

   $$\begin{array}{cc}{x}^3{y}^3& =(xy{)}^3\\ 8& =(2{)}^3\end{array}$$

   Rewrite the expression in the form of a difference of two cubes:

   $$\begin{array}{cc}{x}^3{y}^3-8& =(xy{)}^3-(2{)}^3\end{array}$$

2. Apply the difference of cubes identity.

   Recall the algebraic identity for the difference of two cubes:

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

   Substituting $a=xy$ and $b=2$:

   $$\begin{array}{cc}{x}^3{y}^3-8& =(xy-2)[(xy{)}^2+(xy)(2)+(2{)}^2]\\ & =(xy-2)(x^2{y}^2+2xy+4)\end{array}$$

Solution

1. Factor out the negative sign.

   Factor out $-1$ from all terms of the quadratic polynomial:

   $$\begin{array}{cc}-x^2-23x-60& =-(x^2+23x+60)\end{array}$$

2. Split the middle term.

   For the quadratic expression $x^2+23x+60$, find two numbers whose:

   • Product is equal to the constant term: $60$
   • Sum is equal to the coefficient of the middle term: $23$

   The two required numbers are $20$ and $3$, since:

   $$\begin{array}{cc}20\times 3& =60\\ 20+3& =23\end{array}$$

   Split the middle term $23x$ as $20x+3x$:

   $$\begin{array}{cc}-(x^2+23x+60)& =-(x^2+20x+3x+60)\end{array}$$

3. Factor by grouping.

   Group the terms in pairs and factor out common terms:

   $$\begin{array}{cc}-(x^2+20x+3x+60)& =-[(x^2+20x)+(3x+60)]\\ & =-[x(x+20)+3(x+20)]\\ & =-(x+20)(x+3)\end{array}$$

Solution

1. Identify the product and sum to split the middle term.

   In the quadratic trinomial $a{x}^2+bx+c$:

   $$\begin{array}{cc}a& =2\\ b& =7\\ c& =3\end{array}$$

   We need two numbers whose product is $a\times c=2\times 3=6$ and whose sum is $b=7$. The two numbers are $6$ and $1$, since:

   $$\begin{array}{cc}6\times 1& =6\\ 6+1& =7\end{array}$$

2. Split the middle term and factor by grouping.

   Rewrite $7x$ as $6x+x$:

   $$\begin{array}{cc}2x^2+7x+3& =2x^2+6x+x+3\\ & =(2x^2+6x)+(x+3)\\ & =2x(x+3)+1(x+3)\\ & =(x+3)(2x+1)\end{array}$$

Solution

1. Express terms as squares of monomials.

   Recognize that both terms can be written as perfect squares:

   $$\begin{array}{cc}{x}^4& =(x^2{)}^2\\ 64& =(8{)}^2\end{array}$$

   This matches the sum of two squares $A^2+B^2$ with $A=x^2$ and $B=8$.

2. Complete the square by adding and subtracting $2AB$.

   To form a perfect square trinomial $A^2+2AB+B^2=(A+B{)}^2$, find the required middle term:

   $$\begin{array}{cc}2AB& =2(x^2)(8)\\ & =16x^2\\ & =(4x{)}^2\end{array}$$

   Add and subtract $16x^2$:

   $$\begin{array}{cc}{x}^4+64& =(x^2{)}^2+(8{)}^2\\ & =(x^2{)}^2+2(x^2)(8)+(8{)}^2-16x^2\\ & =[(x^2{)}^2+2(x^2)(8)+(8{)}^2]-(4x{)}^2\\ & =(x^2+8{)}^2-(4x{)}^2\end{array}$$

3. Apply the difference of two squares identity.

   Using $a^2-b^2=(a+b)(a-b)$ with $a=x^2+8$ and $b=4x$:

   $$\begin{array}{cc}(x^2+8{)}^2-(4x{)}^2& =(x^2+8+4x)(x^2+8-4x)\\ & =(x^2+4x+8)(x^2-4x+8)\end{array}$$

4. Verify irreducibility of quadratic factors.

   Calculate the discriminant $D=b^2-4ac$ for each quadratic factor:

   $$\begin{array}{cc}{D}_1& =4^2-4(1)(8)\\ & =16-32\\ & =-16<0\\ D_2& =(-4{)}^2-4(1)(8)\\ & =16-32\\ & =-16<0\end{array}$$

   Since both discriminants are negative, neither quadratic factor can be factored further over the real numbers.

Solution

1. Group the fourth-degree and constant terms.

   Write the terms having square forms together:

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

   Here $A=x^2$ and $B=3$.

2. Determine the term needed to complete the square.

   To complete the square for $(A+B{)}^2=(x^2+3{)}^2$, the required middle term is:

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

   Add and subtract $6x^2$:

   $$\begin{array}{cc}{x}^4+2x^2+9& =(x^2{)}^2+(3{)}^2+2x^2\\ & =(x^2{)}^2+2(x^2)(3)+3^2-6x^2+2x^2\\ & =[(x^2{)}^2+2(x^2)(3)+3^2]-4x^2\\ & =(x^2+3{)}^2-(2x{)}^2\end{array}$$

3. Apply the difference of two squares identity.

   Using $a^2-b^2=(a+b)(a-b)$ with $a=x^2+3$ and $b=2x$:

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

4. Verify irreducibility of quadratic factors.

   Compute the discriminant $D=b^2-4ac$ for each quadratic factor:

   $$\begin{array}{cc}{D}_1& =2^2-4(1)(3)\\ & =4-12\\ & =-8<0\\ D_2& =(-2{)}^2-4(1)(3)\\ & =4-12\\ & =-8<0\end{array}$$

   Since both discriminants are negative, both factors are irreducible over real numbers.

Solution

1. Pair the linear factors by equal sum of constants.

   Compare the sums of the constant terms of the factors:

   $$\begin{array}{cc}3+6& =9\\ 4+5& =9\end{array}$$

   Since the sums are equal, pair $(x+3)$ with $(x+6)$, and $(x+4)$ with $(x+5)$:

   $$\begin{array}{cc}(x+3)(x+4)(x+5)(x+6)-360& =[(x+3)(x+6)][(x+4)(x+5)]-360\end{array}$$

2. Expand each pair.

   $$\begin{array}{cc}(x+3)(x+6)& =x^2+6x+3x+18\\ & =x^2+9x+18\\ (x+4)(x+5)& =x^2+5x+4x+20\\ & =x^2+9x+20\end{array}$$

   Substitute these products back into the expression:

   $$\begin{array}{cc}[(x+3)(x+6)][(x+4)(x+5)]-360& =(x^2+9x+18)(x^2+9x+20)-360\end{array}$$

3. Substitute a temporary variable for the repeated quadratic expression.

   Let $y=x^2+9x$. The expression becomes:

   $$\begin{array}{cc}(y+18)(y+20)-360& =y^2+20y+18y+360-360\\ & =y^2+38y\\ & =y(y+38)\end{array}$$

4. Substitute back $y=x^2+9x$ and factor completely.

   Replace $y$ with $x^2+9x$:

   $$\begin{array}{cc}y(y+38)& =(x^2+9x)(x^2+9x+38)\\ & =x(x+9)(x^2+9x+38)\end{array}$$

5. Verify irreducibility of the quadratic factor.

   Test the discriminant $D=b^2-4ac$ for $x^2+9x+38$:

   $$\begin{array}{cc}D& =9^2-4(1)(38)\\ & =81-152\\ & =-71<0\end{array}$$

   Since $D<0$, the factor $x^2+9x+38$ is irreducible over the real numbers.

Solution

1. Identify the common quadratic expression and substitute.

   Notice that both factors share the common term $x^2+6x$. Let:

   $$\begin{array}{cc}y& =x^2+6x\end{array}$$

   Substitute $y$ into the expression:

   $$\begin{array}{cc}(x^2+6x+3)(x^2+6x-9)+36& =(y+3)(y-9)+36\end{array}$$

2. Expand and simplify the expression in $y$.

   $$\begin{array}{cc}(y+3)(y-9)+36& =y^2-9y+3y-27+36\\ & =y^2-6y+9\end{array}$$

3. Recognize the perfect square trinomial.

   Express as a square of a binomial $(a-b{)}^2=a^2-2ab+b^2$:

   $$\begin{array}{cc}{y}^2-6y+9& =y^2-2(y)(3)+3^2\\ & =(y-3{)}^2\end{array}$$

4. Substitute back $y=x^2+6x$.

   Replace $y$ with $x^2+6x$:

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

5. Check whether the quadratic factor can be factored further.

   Calculate the discriminant $D=b^2-4ac$ for $x^2+6x-3$:

   $$\begin{array}{cc}D& =6^2-4(1)(-3)\\ & =36+12\\ & =48\end{array}$$

   Since $48$ is not a perfect square, $x^2+6x-3$ does not factor into rational linear factors. Thus, the expression is completely factored:

   $$\begin{array}{cc}(x^2+6x+3)(x^2+6x-9)+36& =(x^2+6x-3{)}^2\end{array}$$

Solution

1. Factorize each polynomial completely.

   Factorize the first polynomial by extracting common terms:

   $$\begin{array}{cc}4x^3+12x^2& =4x^2(x+3)\\ & =2^2\cdot x^2\cdot (x+3)\end{array}$$

   Factorize the second polynomial:

   $$\begin{array}{cc}8x^2+16x& =8x(x+2)\\ & =2^3\cdot x\cdot (x+2)\end{array}$$

2. Determine the Highest Common Factor (HCF).

   • Numerical coefficients: $\gcd (2^2,2^3)=2^2=4$
   • Variable factors: $x^{\min (2,1)}=x^1=x$
   • Polynomial binomial factors: $(x+3)$ and $(x+2)$ share no common factor.

   $$\begin{array}{cc}\text{Common factors}& =2^2\cdot x\\ & =4x\end{array}$$

   Therefore:

   $$\begin{array}{cc}\text{HCF}& =4x\end{array}$$

3. Determine the Least Common Multiple (LCM).

   Take the highest power of each distinct factor appearing in either factorization:

   • Highest power of prime $2$: $2^3=8$
   • Highest power of $x$: $x^2$
   • Non-common binomial factors: $(x+3)$ and $(x+2)$

   $$\begin{array}{cc}\text{LCM}& =2^3\cdot x^2\cdot (x+3)(x+2)\\ & =8x^2(x+3)(x+2)\end{array}$$

   (Note on textbook/reference discrepancy: Some references omit one power of $x$ from $4x^2$ and write $8x(x+3)(x+2)$. Mathematically, because the first expression contains $x^2$ and the second contains $x$, the least common multiple must include the highest power $x^2$, giving the exact mathematical LCM $8x^2(x+3)(x+2)$.)