NCERT Solutions for Class 8 Maths Chapter 8 Algebraic Expressions And Identities Ex 8.4

1. Multiply the following binomials.

i. $\left( {2x + 5} \right)$ and $\left( {4x - 3} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {2x + 5} \right) \times \left( {4x - 3} \right) = 2x \times \left( {4x - 3} \right) + 5 \times \left( {4x - 3} \right)$

$\left( {2x + 5} \right) \times \left( {4x - 3} \right) = 8{x^2} - 6x + 20x - 15$

On adding the like terms, we get,

$\left( {2x + 5} \right) \times \left( {4x - 3} \right) = 8{x^2} + 14x - 15$

ii. $\left( {y - 8} \right)$ and $\left( {3y - 4} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {y - 8} \right) \times \left( {3y - 4} \right) = y \times \left( {3y - 4} \right) - 8 \times \left( {3y - 4} \right)$

$\left( {y - 8} \right) \times \left( {3y - 4} \right) = 3{y^2} - 4y - 24y + 32$

On adding the like terms, we get,

$\left( {y - 8} \right) \times \left( {3y - 4} \right) = 3{y^2} - 28y + 32$

iii. $\left( {2.5l - 0.5m} \right)$ and $\left( {2.5l + 0.5m} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {2.5l - 0.5m} \right) \times \left( {2.5l + 0.5m} \right) = 2.5l \times \left( {2.5l + 0.5m} \right) - 0.5m \times \left( {2.5l + 0.5m} \right)$

$\left( {2.5l - 0.5m} \right) \times \left( {2.5l + 0.5m} \right) = 6.25{l^2} + 1.251m - 1.25ml - 0.25{m^2}$

On adding the like terms, we get,

$\left( {2.5l - 0.5m} \right) \times \left( {2.5l + 0.5m} \right) = 6.25{l^2} - 0.25{m^2}$

iv. $\left( {a + 3b} \right)$ and $\left( {x + 5} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {a + 3b} \right) \times \left( {x + 5} \right) = a \times \left( {x + 5} \right) + 3b \times \left( {x + 5} \right)$

$\left( {a + 3b} \right) \times \left( {x + 5} \right) = ax + 5a + 3bx + 15b$

v. $\left( {2pq + 3{q^2}} \right)$ and $\left( {3pq - 2{q^2}} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

\[\left( {2pq + 3{q^2}} \right) \times \left( {3pq - 2{q^2}} \right) = 2pq \times \left( {3pq - 2{q^2}} \right) + 3{q^2} \times \left( {3pq - 2{q^2}} \right)\]

\[\left( {2pq + 3{q^2}} \right) \times \left( {3pq - 2{q^2}} \right) = 6{p^2}{q^2} - 4p{q^3} + 9p{q^3} - 6{q^4}\]

On adding the like terms, we get,

\[\left( {2pq + 3{q^2}} \right) \times \left( {3pq - 2{q^2}} \right) = 6{p^2}{q^2} + 5p{q^3} - 6{q^4}\]

vi. $\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right)$ and $4\left( {{a^2} - \dfrac{2}{3}{b^2}} \right)$

Ans: On multiplying the bracket of the second term and simplifying it, we get,

$4\left( {{a^2} - \dfrac{2}{3}{b^2}} \right) = 4{a^2} - \dfrac{8}{3}{b^2}$

We will multiply the given terms by applying the distributive property and then simplify the equation further. 

\[\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right) \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right) = \dfrac{3}{4}{a^2} \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right) + 3{b^2} \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right)\]

\[\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right) \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right) = 3{a^4} - \dfrac{{24}}{{12}}{a^2}{b^2} + 12{b^2}{a^2} - 8{b^4}\]

\[\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right) \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right) = 3{a^4} - 2{a^2}{b^2} + 12{b^2}{a^2} - 8{b^4}\]

On adding the like terms, we get,

\[\left( {\dfrac{3}{4}{a^2} + 3{b^2}} \right) \times \left( {4{a^2} - \dfrac{8}{3}{b^2}} \right) = 3{a^4} + 10{a^2}{b^2} - 8{b^4}\]

2. Find the product of the following binomials.

i. $\left( {5 - 2x} \right)\left( {3 + x} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {5 - 2x} \right)\left( {3 + x} \right) = 5\left( {3 + x} \right) - 2x\left( {3 + x} \right)$

$\left( {5 - 2x} \right)\left( {3 + x} \right) = 15 + 5x - 6x - 2{x^2}$

On adding the like terms, we get,

$\left( {5 - 2x} \right)\left( {3 + x} \right) = 15 - x - 2{x^2}$

ii. $\left( {x + 7y} \right)\left( {7x - y} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {x + 7y} \right)\left( {7x - y} \right) = x\left( {7x - y} \right) + 7y\left( {7x - y} \right)$

$\left( {x + 7y} \right)\left( {7x - y} \right) = 7{x^2} - xy + 49xy - 7{y^2}$

On adding the like terms, we get,

$\left( {x + 7y} \right)\left( {7x - y} \right) = 7{x^2} + 48xy - 7{y^2}$

iii. $\left( {{a^2} + b} \right)\left( {a + {b^2}} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {{a^2} + b} \right)\left( {a + {b^2}} \right) = {a^2}\left( {a + {b^2}} \right) + b\left( {a + {b^2}} \right)$

$\left( {{a^2} + b} \right)\left( {a + {b^2}} \right) = {a^3} + {a^2}{b^2} + ab + {b^3}$

iv. $\left( {{p^2} - {q^2}} \right)\left( {2p + q} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {{p^2} - {q^2}} \right)\left( {2p + q} \right) = {p^2}\left( {2p + q} \right) - {q^2}\left( {2p + q} \right)$

$\left( {{p^2} - {q^2}} \right)\left( {2p + q} \right) = 2{p^3} + {p^2}q - 2{q^2}p - {q^3}$

3. Simplify the following expressions. 

i. $\left( {{x^2} - 5} \right)\left( {x + 5} \right) + 25$

Ans: We will first multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {{x^2} - 5} \right)\left( {x + 5} \right) + 25 = {x^2}\left( {x + 5} \right) - 5\left( {x + 5} \right) + 25$

$\left( {{x^2} - 5} \right)\left( {x + 5} \right) + 25 = {x^3} + {5x^2} - 5x - 25 + 25$

On adding the like terms, we get,

$\left( {{x^2} - 5} \right)\left( {x + 5} \right) + 25 = {x^3} + {5x^2} - 5x$

ii. $\left( {{a^2} + 5} \right)\left( {{b^3} + 3} \right) + 5$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {{a^2} + 5} \right)\left( {{b^3} + 3} \right) + 5 = {a^2}\left( {{b^3} + 3} \right) + 5\left( {{b^3} + 3} \right) + 5$

$\left( {{a^2} + 5} \right)\left( {{b^3} + 3} \right) + 5 = {a^2}{b^3} + 3{a^2} + 5{b^3} + 15 + 5$

On adding the like terms, we get,

$\left( {{a^2} + 5} \right)\left( {{b^3} + 3} \right) + 5 = {a^2}{b^3} + 3{a^2} + 5{b^3} + 20$

iii. $\left( {t + {s^2}} \right)\left( {{t^2} - s} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

\[\left( {t + {s^2}} \right)\left( {{t^2} - s} \right) = t\left( {{t^2} - s} \right) + {s^2}\left( {{t^2} - s} \right)\]

\[\left( {t + {s^2}} \right)\left( {{t^2} - s} \right) = {t^3} - st + {s^2}{t^2} - {s^3}\]

iv. $\left( {a + b} \right)\left( {c - d} \right) + \left( {a - b} \right)\left( {c + d} \right) + 2\left( {ac + bd} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {a + b} \right)\left( {c - d} \right) + \left( {a - b} \right)\left( {c + d} \right) + 2\left( {ac + bd} \right) = a\left( {c - d} \right) + b\left( {c - d} \right) + a\left( {c + d} \right) - b\left( {c + d} \right) + 2ac + 2bd$

$\left( {a + b} \right)\left( {c - d} \right) + \left( {a - b} \right)\left( {c + d} \right) + 2\left( {ac + bd} \right) = ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd$

On grouping together and adding the like terms, we get,

$\left( {a + b} \right)\left( {c - d} \right) + \left( {a - b} \right)\left( {c + d} \right) + 2\left( {ac + bd} \right) = \left( {ac + ac + 2ac} \right) + \left( {ad - ad} \right) + \left( {bc - bc} \right) + \left( {2bd - bd - bd} \right)$

$\left( {a + b} \right)\left( {c - d} \right) + \left( {a - b} \right)\left( {c + d} \right) + 2\left( {ac + bd} \right) = 4ac$

v. $\left( {x + y} \right)\left( {2x + y} \right) + \left( {x + 2y} \right)\left( {x - y} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {x + y} \right)\left( {2x + y} \right) + \left( {x + 2y} \right)\left( {x - y} \right) = x\left( {2x + y} \right) + y\left( {2x + y} \right) + x\left( {x - y} \right) + 2y\left( {x - y} \right)$

$\left( {x + y} \right)\left( {2x + y} \right) + \left( {x + 2y} \right)\left( {x - y} \right) = 2{x^2} + xy + 2xy + {y^2} + {x^2} - xy + 2yx - 2{y^2}$

On grouping together and adding the like terms, we get,

$\left( {x + y} \right)\left( {2x + y} \right) + \left( {x + 2y} \right)\left( {x - y} \right) = \left( {2{x^2} + {x^2}} \right) + \left( {{y^2} - 2{y^2}} \right) + \left( {xy + 2xy - xy + 2xy} \right)$

$\left( {x + y} \right)\left( {2x + y} \right) + \left( {x + 2y} \right)\left( {x - y} \right) = 3{x^2} - {y^2} + 4xy$

vi. $\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) = x\left( {{x^2} - xy + {y^2}} \right) + y\left( {{x^2} - xy + {y^2}} \right)$

$\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) = {x^3} - {x^2}y + {y^2}x + y{x^2} - x{y^2} + {y^3}$

On grouping together and adding the like terms, we get,

$\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) = {x^3} + {y^3} + \left( {x{y^2} - x{y^2}} \right) + \left( {{x^2}y - {x^2}y} \right)$

$\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) = {x^3} + {y^3}$

vii. $\left( {1.5x - 4y} \right)\left( {1.5x + 4y + 3} \right) - 4.5x + 12y$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {1.5x - 4y} \right)\left( {1.5x + 4y + 3} \right) - 4.5x + 12y = 1.5x\left( {1.5x + 4y + 3} \right) - 4y\left( {1.5x + 4y + 3} \right) - 4.5x + 12y$

$\left( {1.5x - 4y} \right)\left( {1.5x + 4y + 3} \right) - 4.5x + 12y = 2.25{x^2} + 6xy + 4.5x - 6xy - 16{y^2} - 12y - 4.5x + 12y$

On grouping together and adding the like terms, we get,

\[\left( {1.5x - 4y} \right)\left( {1.5x + 4y + 3} \right) - 4.5x + 12y = 2.25{x^2} + \left( {6xy - 6xy} \right) + \left( {4.5x - 4.5x} \right) - 16{y^2} + \left( {12y - 12y} \right)\]

\[\left( {1.5x - 4y} \right)\left( {1.5x + 4y + 3} \right) - 4.5x + 12y = 2.25{x^2} - 16{y^2}\]

viii. $\left( {a + b + c} \right)\left( {a + b - c} \right)$

Ans: We will multiply the given terms by applying the distributive property and then simplify the equation further. 

$\left( {a + b + c} \right)\left( {a + b - c} \right) = a\left( {a + b - c} \right) + b\left( {a + b - c} \right) + c\left( {a + b - c} \right)$

$\left( {a + b + c} \right)\left( {a + b - c} \right) = {a^2} + ab - ac + ab + {b^2} - bc + ac + bc - {c^2}$

On grouping together and adding the like terms, we get,

$\left( {a + b + c} \right)\left( {a + b - c} \right) = {a^2} + {b^2} - {c^2} + \left( {ab + ab} \right) + \left( {bc - bc} \right) + \left( {ac - ac} \right)$

$\left( {a + b + c} \right)\left( {a + b - c} \right) = {a^2} + {b^2} - {c^2} + 2ab$

NCERT Solutions for Class 8 Maths Chapter 8 Algebraic Expressions and Identities Exercise 8.4

Opting for the NCERT solutions for Ex 8.4 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 8.4 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 8 Exercise 8.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 8 Exercise 8.4, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 8 Maths Chapter 8 Exercise 8.4 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

Class 8 Maths Chapter 8: Exercises Breakdown

Other Study Material for CBSE Class 8 Maths Chapter 8

Chapter-wise NCERT Solutions for Class 8 Maths

Important Related Links for CBSE Class 8 Maths