Question

a) simplify $ 3x(4x - 5) + 3 $ and find its value for
i. $ x = 3 $
ii. $ x = 1/2 $
b) Simplify $ a({a^2} + a + 1) + 5 $ and find its value for
i) a=0
ii) a=1
iii)a= -1

Answer 1

Hint: Simplify the given equation using BODMAS and substitute the given value of x in the simplified equation so as to obtain the required value of the given expression.

Complete step-by-step answer:
$3x(4x - 5) + 3$
Simplifying:
$(3x \times 4x) - (3x \times 5) + 3$
$ = 12{x^2} - 15x + 3$ (multiplying like terms)
Substituting the given values of x to find the value
i) $x = 3$

$12{x^2} - 15x + 3 = 12 \times {3^2} - 15(3) + 3$
$ = 12 \times 9 - (15 \times 3) + 3$
$
   = 108 - 45 + 3 \\
   = 108 - 42 \\
   = 66 \\
 $
ii) $x = 1/2$
$12{x^2} - 15x + 3 = 12{(1/2)^2} - 15(1/2) + 3$
 $
   = (12 \times 1/4) - (15/2) + 3 \\
   = 3 - (15/2) + 3 \\
   = 6 - 15/2 \\
 $
Taking LCM, We have:
$
   = (6 \times 2 - 15)/2 \\
   = (12 - 15)/2 \\
   = - 3/2 \\
 $
Therefore, the value of given expression for $x = 3$is 66 and for $x = 1/2$is (-3/2)

b) $a({a^2} + a + 1) + 5$
simplifying:
 $
  a \times {a^2} + a \times a + 1 \times a + 5 \\
   = {a^3} + {a^2} + a + 5 \\
 $
Substituting given value of x to find the value
i) a=0
 ${a^3} + {a^2} + a + 5 = {(0)^3} + {(0)^2} + (0) + 5$
 $ = 5$
ii) a=1
 ${a^3} + {a^2} + a + 5 = {(1)^3} + {(1)^2} + (1) + 5$
 $
   = 1 + 1 + 1 + 5 \\
   = 8 \\
 $

iii) a=-1
${a^3} + {a^2} + a + 5 = {( - 1)^3} + {( - 1)^2} + ( - 1) + 5$
=${( - 1)^{}} + {(1)^{}} + ( - 1) + 5$
  \[
   = - 2 + 6 \\
   = 4 \\
 \]
Therefore, the value of the given expression for a=0, is 5, a= 1 is 8 and for a=-1 is 4.

Note: Even power of (-1) gives positive 1 while odd powers gives negative 1.
Always follow BODMAS for simplification and enclose negative terms in brackets so as to reduce the chance of making calculation mistakes.