Exercise 9.3 Algebraic Expressions and Identities- NCERT Solutions Class 8

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Chapter 9 Ex.9.3 Question 1

Carry out the multiplication of the expressions in each of the following pairs.

(i) \(\quad 4p,\,q + r\)

(ii)\(\quad ab,\,a - b\)

(iii)\(\quad a + b,\,7{a^2}{b^2}\)

(iv)\(\quad {a^2} - 9,\,4a\)

(v)\(\quad pq + qr + rp,\,0\)

Solution

Video Solution

What is known?

Expressions

What is unknown?

Product

Reasoning:

i) By using the distributive law, we can carry out the multiplication term by term.

ii) In multiplication of polynomials with polynomials, we should always look for like terms, if any, and combine them.

Steps:

(i)

\[\begin{align}  &\left( {4p} \right) \times \left( {q + r} \right) \\ &= \left( {4p \times q} \right) + \left( {4p \times r} \right) \\& = 4pq + 4pr\end{align}\]

(ii)

\[\begin{align} & \left( {ab} \right) \times \left( {a - b} \right) \\ &= \left( {ab \times a} \right) + \left[ {ab \times \left( { - b} \right)} \right] \\ &= {a^2}b - a{b^2}\end{align}\]

(iii)

\[\begin{align} & \left( {a + b} \right) \times \left( {7{a^2}{b^2}} \right) \\ &= \left( {a \times 7{a^2}{b^2}} \right) + \left( {b \times 7{a^2}{b^2}} \right) \\ &= 7{a^3}{b^2} + 7{a^2}{b^3}\end{align}\]

(iv)

\[\begin{align}& \left( {{a^2} - 9} \right) \times \left( {4a} \right) \\ &= \left( {{a^2} \times 4a} \right) + \left[ {\left( { - 9} \right) \times \left( {4a} \right)} \right] \\ &= 4{a^3} - 36a \end{align}\]

(v)

\[\begin{align}& \left( {pq + qr + rp} \right) \times 0 \\ &= \left( {pq \times 0} \right) + \left( {qr \times 0} \right) + \left( {rp \times 0} \right) \\ &= 0\end{align}\]

Chapter 9 Ex.9.3 Question 2

Complete the table

---

First expression

Second Expression

Product

(i)

\(a\)

\(b + c + d\)

---

(ii)

\(x + y - 5\)

\(5xy\)

---

(iii)

\(p\)

\(6{p^2} - 7p + 5\)

---

(iv)

\(4{p^2}{q^2}\)

\({p^2} - {q^2}\)

---

(v)

\(a + b + c\)

\(abc\)

---

Solution

Video Solution

What is known?

Expressions

What is unknown?

Product

Steps:

The table can be completed as follows.

---

First expression

Second Expression

Product

(i)

\(a\)

\(b + c + d\)

\(ab + ac + ad\)

(ii)

\(x + y - 5\)

\(5xy\)

\(5{x^2}y + 5x{y^2} - 25xy\)

(iii)

\(p\)

\(6{p^2} - 7p + 5\)

\(6{p^3} - 7{p^2} + 5p\)

(iv)

\(4{p^2}{q^2}\)

\({p^2} - {q^2}\)

\(4{p^4}{q^2} - 4{p^2}{q^4}\)

(v)

\(a + b + c\)

\(abc\)

\({a^2}bc + a{b^2}c + ab{c^2}\)

Chapter 9 Ex.9.3 Question 3

Find the product.

(i) \(\begin{align}\left( {{a^2}} \right) \times \left( {2{a^{22}}} \right) \times \left( {4{a^{26}}} \right) \end{align}\)

(ii) \(\begin{align}\left( {\frac{2}{3}xy} \right) \times \left( {\frac{{ - 9}}{{10}}{x^2}{y^2}} \right) \end{align}\)

(iii) \(\begin{align}\left( { - \frac{{10}}{3}p{q^3}} \right) \times \left( {\frac{6}{5}{p^3}q} \right) \end{align}\)

(iv) \(\begin{align} x \times {x^{2}} \times {x^3} \times {x^4} \end{align}\)

Solution

Video Solution

What is known?

Expressions

What is unknown?

Simplification

Steps:

(i)

\[\begin{align} &\left( {{a^2}} \right) \times \left( {2{a^{22}}} \right) \times \left( {4{a^{26}}} \right) \\&= 2 \times 4 \times {a^2} \times {a^{22}} \times {a^{26}} \\&= 8{a^{50}}\end{align}\]

(ii)

\[\begin{align} &\left( {\frac{2}{3}xy} \right) \times \left( {\frac{{ - 9}}{{10}}{x^2}{y^2}} \right) \\&= \!\! \left( {\frac{\not{2}}{3}} \right) \! \! \times \!\!  \left( {\frac{{ - 9}}{{\not{\!\!10^5}}}} \right) \! \times  \! x \times \! y \! \times {x^2} \! \times \! {y^2} \\&= \frac{{ - 3}}{5}{x^3}{y^3}\end{align}\]

(iii)

\[\begin{align} & \left( \frac{-10}{3}p{{q}^{3}} \right)\times \left( \frac{6}{5}{{p}^{3}}q \right)\\&=\left(\frac{-1{{{\not\!0}}^{2}}}{3} \right)\times \left( \frac{6}{{\not\!5}} \right)\times p{{q}^{3}}\times{{p}^{3}}q\\&=-4{{p}^{4}}{{q}^{4}}\end{align}\]

(iv)

\[x \times {x^2} \times {x^3} \times {x^4} = {x^{10}}\]

Chapter 9 Ex.9.3 Question 4

(a) Simplify \(3x\left( {4x - 5} \right) + 3\) and find its values for

(i) \(\;x = 3\) 

(ii)\(\begin{align}\;x = \frac{1}{2}\end{align}\)

(b) \(a\left( {{a^2} + a + 1} \right) + 5\) and find its values for

(i) \(\;a = 0\)

(ii)\(\;a = 1\)

(iii) \(a =- 1\)

Solution

Video Solution

What is known?

Expression with their corresponding values.

What is unknown?

Simplification and its result with their corresponding values

Steps:

(a) \(3x\left( {4x - 5} \right) + 3 = 12{x^2} - 15x + 3\)

(i)

\[\begin{align}{\text{For }}x &= {\rm{ }}3,\\& =12{x^2} - 15x + 3\\& = 12{\left( 3 \right)^2} - 15\left( 3 \right) + 3\\&= 108 - 45+ 3\\&= 66\end{align}\]

(ii)

For \(x=\frac{1}{2}\) ,

\[\begin{align}  & 12{{x}^{2}}-15x+3 \\ & =12{{\left( \frac{1}{2} \right)}^{2}}-15\left( \frac{1}{2} \right)+3 \\ & =\,\not\!\!\!{{12}^{3}}\times \frac{1}{{\not\!4}}-\frac{15}{2}+3 \\ & =3-\frac{15}{2}+3 \\ & =6-\frac{15}{2} \\ & =\frac{12-15}{2}=\frac{-3}{2} \\ \end{align}\]

(b)

\(\begin{align} a\left( {{a^2} + a +1} \right) + 5\\ = {a^3} + {a^2} + a + 5\end{align} \)

(i)

For \(a = 0\) ,

\[\begin{align}&{a^3} + {a^2} + a + 5 \\&= 0 + 0 + 0 + 5 \\&= 5 \end{align}\]

(ii)

For \(a = 1\),

\[\begin{align}& {a^3} + {a^2} + a + 5 \\&= {\rm{ }}{{\left( 1 \right)}^3} + {\rm{ }}{{\left( 1 \right)}^2} + 1+ 5\\& = 1+1+ 1 + 5\\&=8 \end{align}\]

(iii)

For \(a = - 1\) ,

\[\begin{align}&{a^3} + {a^2} + a + 5\\&= {{\left( { - 1} \right)}^3} + {{\left( { - 1} \right)}^2} + \left( { - 1} \right) + 5\\&= - 1 + 1 - 1 + 5\\&= 4\end{align}\]

Chapter 9 Ex.9.3 Question 5

(a) Add: \(p\left( {p - q} \right),q\left( {q - r} \right)\) and \(r\left( {r - p} \right)\)

(b) Add: \(2x\left( {z - x - y} \right)\) and \(2y\left( {z - y - x} \right)\)

(c) Subtract: \(3l\left( {l - 4m + 5n} \right)\) from \(4l\left( {10n - 3m + 2l} \right)\)

(d) Subtract: \(3a\left( {a + b + c} \right) - 2b\left( {a - b + c} \right)\) from \(4c\left( { - a + b + c} \right)\)

Solution

Video Solution

What is known?

Expressions

What is unknown?

Simplification

Reasoning:

Addition and Subtraction takes place between like terms.

Steps:

(a)       

First expression

\(=p\left( {p{\rm{ }} - {\rm{ }}q} \right) = {p^2} - pq\)

Second expression

\(=q\left( {q - r} \right) = {q^2} - qr\)

Third expression 

\(=r\left( {r - p} \right) = {r^2} - pr\)

Adding the three expressions, we obtain

\[\frac{\begin{align}{p^2} - pq&\\+\;\; &{q^2} - qr\\+& \;\;{r^2} - pq\end{align}}{{{p^2} - pq + {q^2} - qr + {r^2} - pq}}\]

Therefore, the sum of the given expressions is \({p^2} + {q^2} + {r^2} - pq - qr - rp.\)

(b)     

First expression

\[\begin{align}&=2x\left( {z - x - y} \right) \\ &= 2xz - 2{x^2} - 2xy\end{align}\]

Second expression

\(\begin{align}&=2y\left( {z - y - x} \right) \\ &= 2yz - 2{y^2} - 2yx\end{align} \)

Adding the two expressions, we obtain

\begin{align}\frac{\begin{align}& 2xz-2{{x}^{2}}-2xy \\& +\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2xy+2yz-2{{y}^{2}} \\\end{align}}{2xz-2{{x}^{2}}-4xy+2yz-2{{y}^{2}}}\end{align}

Therefore, the sum of the given expressions is \( - 2{x^2} - 2{y^2} - 4xy + 2yz + 2zx.\)

(c)

\[\begin{align}3l\left( {l \!- \!4m \!+ \!5n} \right) &= 3{l^2} \!-\! 12lm \!+ \!15ln \\4l\left( {10n\! -\! 3m \!+\! 2l} \right) &=40ln \!-\! 12lm \!+\! 8{l^2} \end{align}\]

Subtracting these expressions, we obtain

\[\frac{\begin{align}40ln - 12lm + 8{l^2}\\15l - 12lm + 3{l^2}\\\left(-\right)\,\,\,\,\,\,\,\left(+\right)\,\,\,\,\,\,\,\,\,\left(-\right)\end{align}}{{ + 25ln\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 5{l^2}}}\]

Therefore, the result is \(5{l^2} + 25ln.\)

(d)

\[\begin{align}&3a\left( {a + b + c} \right) - 2b\left( {a - b + c} \right)\\&=\begin{bmatrix} 3{a^2} + 3ab + 3ac- \\ 2ba + 2{b^2} - 2bc \end{bmatrix}\\&= \begin{bmatrix} 3{a^{2}} + 2{b^2} +\\ ab + 3ac - 2bc\end{bmatrix} \\&4c\left( \! - a + b + c \right) \!= - 4ac + 4bc  + 4{c^2} \end{align}\]

Subtracting these expressions, we obtain

\(\dfrac{\begin{align}& -4ac+4bc+4{{c}^{2}} \\& \,\,\,\,\,3ac-2bc\,\,\,\,\,\,\,\,\,\,\,+3{{a}^{2}}+2{{b}^{2}}+ab \\& \left( - \right)\,\,\,\,\,\,\,\,\,\,\,\left( + \right)\,\,\,\,\,\,\,\,\,\,\,\left( - \right)\,\,\,\,\,\,\,\,\left( -\right)\,\,\,\,\,\left( - \right) \\\end{align}}{-7ac+6bc+4{{c}^{2}}\!-\!3{{a}^{2}}\!-2{{b}^{2}}\!-ab}\)

Therefore, the result is \( - 3{a^2} - 2{b^2} +\! 4{c^2} - ab \!+ 6bc - 7ac\)

  
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