Polynomials Test - 29

$$ {208}^{2} = {(200 + 8)}^{2} $$

It is the form of $$ {(a+b)}^{2} $$, where $$ a = 200, b = 8 $$

Applying

the formula $$ { (a+b) }^{ 2 } = {a}^{2}  + { b }^{ 2 } + 2ab $$
$$ {208}^{2} = {(200 + 8)}^{2} = {200}^{2}  + { 8 }^{ 2 } + 2\times 200 \times 8 = 40000 + 64 + 3200 = 43264 $$

$${ (a+b) }^{ 2 }\quad =\quad a^{ 2 }+b^{ 2 }+2ab\\ a=9,\quad b=0.4\\ =\quad 9^{ 2 }+0.4^{ 2 }+2*9*0.4\\ =\quad 88.36$$

$$ {188}^{2} = {(200 - 12)}^{2} $$

It is the form of $$ {(a - b)}^{2} $$, where $$ a = 200, b = 12 $$

Applying the formula $$ { (a - b) }^{ 2 } = {a}^{2}  + { b }^{ 2 } - 2ab $$
$$  {188}^{2} = {(200 - 12)}^{2} = {200}^{2}  + { 12 }^{ 2 } - 2\times 200 \times 12 = 40000 + 144 - 4800 = 35344 $$

Given: $$ {92}^{2} = {(90 + 2)}^{2} $$

It is in the form of $$ {(a+b)}^{2} $$, where $$ a = 90, b = 2 $$.

Applying the formula $$ { (a+b) }^{ 2 } = {a}^{2}  + { b }^{ 2 } + 2ab $$ we get,

$$ 203 \times 197 = (200 + 3) \times (200 - 3) $$

Applying the formula $$ (a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 } $$, where $$ a =

200 , b = 3 $$

$$ 203 \times 197 = (200 + 3) \times (200 - 3) = { 200 }^{ 2 }-{ 3 }^{ 2 } = 40,000 - 9 = 39991 $$

$$ 20.8 \times 19.2 = (20+ 0.8) \times (20 - 0.8) $$

Applying the formula $$ (a+b)(a-b) = { a }^{ 2 }-{ b }^{ 2 } $$, where $$ a =

20 , b = 0.8 $$

$$ 20.8 \times 19.2 = (20+ 0.8) \times (20 - 0.8) = { 20 }^{ 2 }-{ 0.8 }^{ 2 } = 400 - 0.64

= 399.36 $$

$$ {391}^{2} = {(400 - 9)}^{2} $$

It is the form of $$ {(a - b)}^{2} $$, where $$ a = 400, b = 9 $$

Applying the formula $$ { (a - b) }^{ 2 } = {a}^{2}  + { b }^{ 2 } - 2ab

$$
$$ {391}^{2} = {(400 - 9)}^{2} = {400}^{2}  + { 9 }^{ 2 } -

2\times 400 \times 9 = 1,60,000 + 81 - 7200 = 152881 $$

$$607$$ can be written as $$600+7$$

$$\therefore {607}^{2} = {(600 + 7)}^{2} $$
It is the form of $$ {(a+b)}^{2} $$, where $$ a = 600, b = 7 $$
Applying the formula $$ { (a+b) }^{ 2 } = {a}^{2}  + { b }^{ 2 } + 2ab $$
$$  {607}^{2} = {(600 + 7)}^{2} = {600}^{2}  + { 7 }^{ 2 } +2\times 600 \times 7 = 360000 + 49 + 8400 = 368449 $$

If $$1$$ is the zero of the polynomial $$f(x) =a^2x^2 - 3ax +3x - 1$$
Then, $$f(x)=0$$
$$=>a^2(1)^2-3a(1)+3(1)-1=0$$
$$=>a^2-3a+3-1=0$$
$$=>a^2-3a+2=0$$
$$=>a^2-2a-a+2=0$$
$$=>a(a-2)-1(a-2)=0$$
$$=>(a-2)(a-1)=0$$
$$a=2$$
Hence, option $$B$$ is correct.

If $$2$$  and $$ 3 $$ are the zeros of the quadratic polynomial then
$$\left( x-2  \right) \left( x-3  \right)$$ are the factors of the polynomial
Thus,