Exercise 4.1 — Class 9 Mathematics Solutions

Solution

Solution

1. Prime and monomial factorization of each term.

   Expand each term into its prime and variable factors:

   $$\begin{array}{cc}6x& =2\times 3\times x\\ 12& =2\times 2\times 3\end{array}$$

   The common prime factors shared by both terms are $2$ and $3$. Hence, the greatest common factor (GCF) is:

   $$\begin{array}{cc}\text{GCF}& =2\times 3\\ & =6\end{array}$$

2. Factor out the greatest common factor.

   Express each term as a product with $6$ and apply the distributive property:

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

Solution

1. Prime and monomial factorization of each term.

   Expand each term into its prime and variable factors:

   $$\begin{array}{cc}15y^2& =3\times 5\times y\times y\\ 20y& =2\times 2\times 5\times y\end{array}$$

   The common factors shared by both terms are $5$ and $y$. Thus, the greatest common factor (GCF) is:

   $$\begin{array}{cc}\text{GCF}& =5\times y\\ & =5y\end{array}$$

2. Factor out the greatest common factor.

   Rewrite each term with $5y$ factored out:

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

Solution

1. Prime and monomial factorization of each term.

   Expand each term into its prime and variable factors, noting the shared negative sign:

   $$\begin{array}{cc}-12x^2& =-1\times 2\times 2\times 3\times x\times x\\ -3x& =-1\times 3\times x\end{array}$$

   The common factors shared by both terms are $-1$, $3$, and $x$. Thus, the greatest common factor (GCF) is:

   $$\begin{array}{cc}\text{GCF}& =-1\times 3\times x\\ & =-3x\end{array}$$

2. Factor out the greatest common factor.

   Factor out $-3x$ from each term, leaving positive terms inside the parentheses:

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

Solution

1. Prime and monomial factorization of each term.

   Expand each term into prime and variable factors:

   $$\begin{array}{cc}4a^{2}b& =2\times 2\times a\times a\times b\\ 8a{b}^2& =2\times 2\times 2\times a\times b\times b\end{array}$$

   The common factors are two $2$'s, one $a$, and one $b$. Hence, the greatest common factor (GCF) is:

   $$\begin{array}{cc}\text{GCF}& =2\times 2\times a\times b\\ & =4ab\end{array}$$

2. Factor out the greatest common factor.

   Express each term as a product with $4ab$ and apply the distributive property:

   $$\begin{array}{cc}4a^{2}b+8a{b}^2& =4ab(a)+4ab(2b)\\ & =4ab(a+2b)\end{array}$$

Solution

1. Prime and monomial factorization of each term.

   Expand each term into its constituent factors:

   $$\begin{array}{cc}xy& =x\times y\\ -3x^2& =-1\times 3\times x\times x\\ 2x& =2\times x\end{array}$$

   Comparing the three terms, the only factor common to all three is $x$. Thus, the greatest common factor (GCF) is:

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

2. Factor out the greatest common factor.

   Rewrite each term as a multiple of $x$ and factor it out:

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

Solution

1. Prime and monomial factorization of each term.

   Expand each of the three terms into prime and variable factors:

   $$\begin{array}{cc}3a^{2}b& =3\times a\times a\times b\\ -9a{b}^2& =-1\times 3\times 3\times a\times b\times b\\ 15ab& =3\times 5\times a\times b\end{array}$$

   The common factors present in all three terms are $3$, $a$, and $b$. Therefore, the greatest common factor (GCF) is:

   $$\begin{array}{cc}\text{GCF}& =3\times a\times b\\ & =3ab\end{array}$$

2. Factor out the greatest common factor.

   Express each term as a product with $3ab$ and factor it out:

   $$\begin{array}{cc}3a^{2}b-9a{b}^2+15ab& =3ab(a)-3ab(3b)+3ab(5)\\ & =3ab(a-3b+5)\end{array}$$

Solution

Solution

1. Identify the common numerical factor.

   Examine the numerical coefficients $5$ and $15$:

   • The greatest common divisor (GCD / HCF) of $5$ and $15$ is $5$.

2. Factorize algebraically.

   Factor out the common factor $5$ from each term:

   $$\begin{array}{cc}5x+15& =5(x)+5(3)\\ & =5(x+3)\end{array}$$

3. Pictorial / Algebra Tile Area Model Representation.

   In the algebra tile area model:

   • The expression represents a rectangle with height $5$ and width $x+3$.
   • Here, the factor $(x+3)$ is repeated $5$ times vertically.
   • The width $(x+3)$ is partitioned into $x+1+1+1$.
   • The height $5$ is partitioned into $5$ individual unit rows of height $1$.
   • The composite rectangle consists of:
     • $5$ rectangular tiles each of dimensions $x\times 1$, contributing an area of $5x$.
     • $15$ unit square tiles each of dimensions $1\times 1$, contributing an area of $15$.
   • Summing the areas verifies the algebraic identity:

   $$\begin{array}{cc}\text{Total Area}& =\text{Height}\times \text{Width}\\ & =5(x+3)\\ & =5x+15\end{array}$$

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$x^2+4x+3$$

   Comparing with $a{x}^2+bx+c$:

   • Leading coefficient: $a=1$
   • Middle coefficient: $b=4$
   • Constant term: $c=3$

   Highlighted Side Calculations:

   • Target Product: $P=a\times c=1\times 3=3$
   • Target Sum: $S=b=4$
   • Factor Pair: We find two integers whose product is $3$ and sum is $4$:

   $$\begin{array}{cc}3\times 1& =3\\ 3+1& =4\end{array}$$

   The factor pair is $3$ and $1$.

2. Split the middle term and factor by grouping.

   Split $4x$ into $3x+x$:

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

3. Pictorial / Algebra Tiles Area Model Representation.

   In the algebra tile area model, the expression forms a composite rectangle with side lengths $(x+3)$ and $(x+1)$:

   • Horizontal dimension: $x+3$, divided into one segment of length $x$ and three unit segments of length $1$.
   • Vertical dimension: $x+1$, divided into one segment of height $x$ and one segment of height $1$.
   • The grid consists of:
     • $1$ large square tile of dimensions $x\times x$ with area $x^2$.
     • $3$ rectangular strip tiles along the top of dimensions $1\times x$, each having area $x$ (sum = $3x$).
     • $1$ rectangular strip tile along the left side of dimensions $x\times 1$ with area $x$.
     • $3$ unit square tiles in the corner of dimensions $1\times 1$, each having area $1$ (sum = $3$).
   • Summing all the regions:

   $$\begin{array}{cc}\text{Total Area}& =x^2+3x+x+3\\ & =x^2+4x+3\\ & =(x+3)(x+1)\end{array}$$

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$x^2+6x+8$$

   Comparing with $a{x}^2+bx+c$:

   • Leading coefficient: $a=1$
   • Middle coefficient: $b=6$
   • Constant term: $c=8$

   Highlighted Side Calculations:

   • Target Product: $P=a\times c=1\times 8=8$
   • Target Sum: $S=b=6$
   • Factor Pair: We find two integers whose product is $8$ and sum is $6$:

   $$\begin{array}{cc}4\times 2& =8\\ 4+2& =6\end{array}$$

   The factor pair is $4$ and $2$.

2. Split the middle term and factor by grouping.

   Split $6x$ into $4x+2x$:

   $$\begin{array}{cc}{x}^2+6x+8& =x^2+4x+2x+8\\ & =x(x+4)+2(x+4)\\ & =(x+4)(x+2)\end{array}$$

3. Pictorial / Algebra Tiles Area Model Representation.

   Using algebra tiles, the expression is arranged into a solid rectangle of dimensions $(x+4)\times (x+2)$:

   • Horizontal dimension: $x+4$, partitioned into $x+1+1+1+1$.
   • Vertical dimension: $x+2$, partitioned into $x+1+1$.
   • The grid consists of:
     • $1$ large square tile of dimensions $x\times x$ with area $x^2$.
     • $4$ rectangular strip tiles of dimensions $1\times x$ along row 1, each with area $x$ (sum = $4x$).
     • $2$ rectangular strip tiles of dimensions $x\times 1$ along column 1, each with area $x$ (sum = $2x$).
     • $8$ unit square tiles of dimensions $1\times 1$ arranged in a $2\times 4$ corner block, each with area $1$ (sum = $8$).
   • The sum of all tile areas equals the total rectangle area:

   $$\begin{array}{cc}\text{Total Area}& =(x+4)(x+2)\\ & =x^2+4x+2x+8\\ & =x^2+6x+8\end{array}$$

Solution

1. Recognize the perfect square trinomial structure.

   Notice that:

   • First term is a square: $x^2=(x{)}^2$
   • Last term is a square: $4=(2{)}^2$
   • Middle term is twice the product of the terms: $4x=2(x)(2)$

   Highlighted Side Calculations: Using the standard identity $(a+b{)}^2=a^2+2ab+b^2$ with $a=x$ and $b=2$:

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

   Alternatively, splitting the middle term with product $4$ and sum $4$ gives the factor pair $2$ and $2$.

2. Factorize by grouping or identity.

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

3. Pictorial / Algebra Tiles Area Model Representation.

   In the algebra tile model, this expression forms a perfect geometric square of side length $(x+2)$:

   • Horizontal dimension: $x+2$, partitioned into $x+1+1$.
   • Vertical dimension: $x+2$, partitioned into $x+1+1$.
   • The grid consists of:
     • $1$ large square tile of dimensions $x\times x$ with area $x^2$ in the top-left.
     • $2$ rectangular strip tiles of dimensions $1\times x$ along the top row, each with area $x$ (sum = $2x$).
     • $2$ rectangular strip tiles of dimensions $x\times 1$ along the first column, each with area $x$ (sum = $2x$).
     • $4$ unit square tiles of dimensions $1\times 1$ filling the $2\times 2$ bottom-right corner, each with area $1$ (sum = $4$).
   • Combining all tiles produces a complete symmetric square of area:

   $$\begin{array}{cc}\text{Total Area}& =(x+2{)}^2\\ & =x^2+2x+2x+4\\ & =x^2+4x+4\end{array}$$

Solution

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$x^2+x-12$$

   Comparing with the standard quadratic form $a{x}^2+bx+c$:

   • Leading coefficient: $a=1$
   • Middle coefficient: $b=1$
   • Constant term: $c=-12$

   Highlighted Side Calculation (Product & Sum):

   Quantity	Formula / Condition	Value
   Product	$a\times c=1\times (-12)$	$-12$
   Sum	$b$	$1$
   Two factors $(p,q)$	$p\times q=ac$ and $p+q=b$	$4$ and $-3$

   We find two integers $p$ and $q$ such that their product is $-12$ and their sum is $1$:

   $$\begin{array}{cc}p\times q& =4\times (-3)=-12\\ p+q& =4+(-3)=1\end{array}$$

   The factor pair is $4$ and $-3$, which splits the middle term $+x$ into $+4x-3x$.

2. Split the middle term and factor by grouping.

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

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$x^2+7x+10$$

   Comparing with the standard quadratic form $a{x}^2+bx+c$:

   • Leading coefficient: $a=1$
   • Middle coefficient: $b=7$
   • Constant term: $c=10$

   Highlighted Side Calculation (Product & Sum):

   Quantity	Formula / Condition	Value
   Product	$a\times c=1\times 10$	$10$
   Sum	$b$	$7$
   Two factors $(p,q)$	$p\times q=ac$ and $p+q=b$	$5$ and $2$

   We find two integers $p$ and $q$ such that their product is $10$ and their sum is $7$:

   $$\begin{array}{cc}p\times q& =5\times 2=10\\ p+q& =5+2=7\end{array}$$

   The factor pair is $5$ and $2$, which splits the middle term $+7x$ into $+5x+2x$.

2. Split the middle term and factor by grouping.

   $$\begin{array}{cc}{x}^2+7x+10& =x^2+5x+2x+10\\ & =x(x+5)+2(x+5)\\ & =(x+5)(x+2)\end{array}$$

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$x^2-6x+8$$

   Comparing with the standard quadratic form $a{x}^2+bx+c$:

   • Leading coefficient: $a=1$
   • Middle coefficient: $b=-6$
   • Constant term: $c=8$

   Highlighted Side Calculation (Product & Sum):

   Quantity	Formula / Condition	Value
   Product	$a\times c=1\times 8$	$8$
   Sum	$b$	$-6$
   Two factors $(p,q)$	$p\times q=ac$ and $p+q=b$	$-4$ and $-2$

   We find two integers $p$ and $q$ such that their product is $8$ and their sum is $-6$:

   $$\begin{array}{cc}p\times q& =(-4)\times (-2)=8\\ p+q& =(-4)+(-2)=-6\end{array}$$

   The factor pair is $-4$ and $-2$, which splits the middle term $-6x$ into $-4x-2x$.

2. Split the middle term and factor by grouping.

   $$\begin{array}{cc}{x}^2-6x+8& =x^2-4x-2x+8\\ & =x(x-4)-2(x-4)\\ & =(x-4)(x-2)\end{array}$$

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$x^2-x-56$$

   Comparing with the standard quadratic form $a{x}^2+bx+c$:

   • Leading coefficient: $a=1$
   • Middle coefficient: $b=-1$
   • Constant term: $c=-56$

   Highlighted Side Calculation (Product & Sum):

   Quantity	Formula / Condition	Value
   Product	$a\times c=1\times (-56)$	$-56$
   Sum	$b$	$-1$
   Two factors $(p,q)$	$p\times q=ac$ and $p+q=b$	$-8$ and $7$

   We find two integers $p$ and $q$ such that their product is $-56$ and their sum is $-1$:

   $$\begin{array}{cc}p\times q& =(-8)\times 7=-56\\ p+q& =(-8)+7=-1\end{array}$$

   The factor pair is $-8$ and $7$, which splits the middle term $-x$ into $-8x+7x$.

2. Split the middle term and factor by grouping.

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

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$x^2-10x-24$$

   Comparing with the standard quadratic form $a{x}^2+bx+c$:

   • Leading coefficient: $a=1$
   • Middle coefficient: $b=-10$
   • Constant term: $c=-24$

   Highlighted Side Calculation (Product & Sum):

   Quantity	Formula / Condition	Value
   Product	$a\times c=1\times (-24)$	$-24$
   Sum	$b$	$-10$
   Two factors $(p,q)$	$p\times q=ac$ and $p+q=b$	$-12$ and $2$

   We find two integers $p$ and $q$ such that their product is $-24$ and their sum is $-10$:

   $$\begin{array}{cc}p\times q& =(-12)\times 2=-24\\ p+q& =(-12)+2=-10\end{array}$$

   The factor pair is $-12$ and $2$, which splits the middle term $-10x$ into $-12x+2x$.

2. Split the middle term and factor by grouping.

   $$\begin{array}{cc}{x}^2-10x-24& =x^2-12x+2x-24\\ & =x(x-12)+2(x-12)\\ & =(x-12)(x+2)\end{array}$$

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$y^2+4y-12$$

   Comparing with the standard quadratic form $a{y}^2+by+c$:

   • Leading coefficient: $a=1$
   • Middle coefficient: $b=4$
   • Constant term: $c=-12$

   Highlighted Side Calculation (Product & Sum):

   Quantity	Formula / Condition	Value
   Product	$a\times c=1\times (-12)$	$-12$
   Sum	$b$	$4$
   Two factors $(p,q)$	$p\times q=ac$ and $p+q=b$	$6$ and $-2$

   We find two integers $p$ and $q$ such that their product is $-12$ and their sum is $4$:

   $$\begin{array}{cc}p\times q& =6\times (-2)=-12\\ p+q& =6+(-2)=4\end{array}$$

   The factor pair is $6$ and $-2$, which splits the middle term $+4y$ into $+6y-2y$.

2. Split the middle term and factor by grouping.

   $$\begin{array}{cc}{y}^2+4y-12& =y^2+6y-2y-12\\ & =y(y+6)-2(y+6)\\ & =(y+6)(y-2)\end{array}$$

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$y^2+13y+36$$

   Comparing with the standard quadratic form $a{y}^2+by+c$:

   • Leading coefficient: $a=1$
   • Middle coefficient: $b=13$
   • Constant term: $c=36$

   Highlighted Side Calculation (Product & Sum):

   Quantity	Formula / Condition	Value
   Product	$a\times c=1\times 36$	$36$
   Sum	$b$	$13$
   Two factors $(p,q)$	$p\times q=ac$ and $p+q=b$	$9$ and $4$

   We find two integers $p$ and $q$ such that their product is $36$ and their sum is $13$:

   $$\begin{array}{cc}p\times q& =9\times 4=36\\ p+q& =9+4=13\end{array}$$

   The factor pair is $9$ and $4$, which splits the middle term $+13y$ into $+9y+4y$.

2. Split the middle term and factor by grouping.

   $$\begin{array}{cc}{y}^2+13y+36& =y^2+9y+4y+36\\ & =y(y+9)+4(y+9)\\ & =(y+9)(y+4)\end{array}$$

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$x^2-x-2$$

   Comparing with the standard quadratic form $a{x}^2+bx+c$:

   • Leading coefficient: $a=1$
   • Middle coefficient: $b=-1$
   • Constant term: $c=-2$

   Highlighted Side Calculation (Product & Sum):

   Quantity	Formula / Condition	Value
   Product	$a\times c=1\times (-2)$	$-2$
   Sum	$b$	$-1$
   Two factors $(p,q)$	$p\times q=ac$ and $p+q=b$	$-2$ and $1$

   We find two integers $p$ and $q$ such that their product is $-2$ and their sum is $-1$:

   $$\begin{array}{cc}p\times q& =(-2)\times 1=-2\\ p+q& =(-2)+1=-1\end{array}$$

   The factor pair is $-2$ and $1$, which splits the middle term $-x$ into $-2x+x$.

2. Split the middle term and factor by grouping.

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

Solution

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$2x^2+7x+3$$

   Comparing with standard form $a{x}^2+bx+c$:

   • Leading coefficient: $a=2$
   • Middle coefficient: $b=7$
   • Constant term: $c=3$

   Highlighted Side Calculations:

   • Target Product: $P=a\times c=2\times 3=6$
   • Target Sum: $S=b=7$
   • Factor Pair: We find two integers whose product is $6$ and sum is $7$. Since both product and sum are positive, both factors are positive:

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

   The factor pair is $6$ and $1$.

2. Split the middle term and factor by grouping.

   Split $7x$ into $6x+x$:

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

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$2x^2+11x+15$$

   Comparing with standard form $a{x}^2+bx+c$:

   • Leading coefficient: $a=2$
   • Middle coefficient: $b=11$
   • Constant term: $c=15$

   Highlighted Side Calculations:

   • Target Product: $P=a\times c=2\times 15=30$
   • Target Sum: $S=b=11$
   • Factor Pair: We find two integers whose product is $30$ and sum is $11$. Since both product and sum are positive, both factors are positive:

   $$\begin{array}{cc}6\times 5& =30\\ 6+5& =11\end{array}$$

   The factor pair is $6$ and $5$.

2. Split the middle term and factor by grouping.

   Split $11x$ into $6x+5x$:

   $$\begin{array}{cc}2x^2+11x+15& =2x^2+6x+5x+15\\ & =2x(x+3)+5(x+3)\\ & =(x+3)(2x+5)\end{array}$$

Solution

1. Identify the product, sum, and factor pair.

   Given quadratic expression:

   $$4x^2+13x+3$$

   Comparing with standard form $a{x}^2+bx+c$:

   • Leading coefficient: $a=4$
   • Middle coefficient: $b=13$
   • Constant term: $c=3$

   Highlighted Side Calculations:

   • Target Product: $P=a\times c=4\times 3=12$
   • Target Sum: $S=b=13$
   • Factor Pair: We find two integers whose product is $12$ and sum is $13$. Since both product and sum are positive, both factors are positive:

   $$\begin{array}{cc}12\times 1& =12\\ 12+1& =13\end{array}$$

   The factor pair is $12$ and $1$.

2. Split the middle term and factor by grouping.

   Split $13x$ into $12x+x$:

   $$\begin{array}{cc}4x^2+13x+3& =4x^2+12x+x+3\\ & =4x(x+3)+1(x+3)\\ & =(x+3)(4x+1)\end{array}$$