# Ex.9.3 Q4 Algebraic Expressions and Identities - NCERT Maths Class 8

## Question

(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) $$a =- 1$$

Video Solution
Algebraic Expressions & Identities
Ex 9.3 | Question 4

## Text 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}

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school