Maths M.C.Q. [1 Marks Each]

Apolynomialis anexpressionconsisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. the variable is $x$ but in $\frac{7}{\text{x}}$ power of xis not a whole number. this is not a polynomial.

Given,
$\Rightarrow a+b=10 \Rightarrow a b=16 $
$ \Rightarrow(a+b)^2=a^2+b^2+2 a b $
$ \Rightarrow 10^2=a^2+b^2+2(16) $
$ \therefore a^2+b^2=68 $
$ \Rightarrow a^2+b^2+a b=68+16=84 $
$ \Rightarrow a^2+b^2-a b=68-16=52$

The maximum no. of terms in a polynomial of degree $10$ is a polynomial that can have terms with powers of $x$ as $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ and $10$. there are $11$ such terms that can be possible with these powers of $x$ and $a$ real coefficient.

Let the polynomial to be added be $p$
$\therefore x^3+3 x-8+p=3 x^3+x^2+6 $
$ \therefore p=3 x^3+x^2+6-x^3-3 x+8 $
$ \therefore p=2 x^3+x^2-3 x+14$

$(4x^3 + 2x^2 + 4x + 4) × 2$
$= 8x^3 + 4x^2 + 8x + 8$

Since, $(x^2 + 2x - 3) - (-2x^2 + x + 1)$
$= x^2 + 2x - 3 + 2x^2 - x - 1$
$= 3x^2 + x - 4$
So, $-2x^2 + x + 1$ is less than $x^2 + 2x - 3$ by $3x^2 + x - 4.$
Hence, the correct alternative is option $(b).$

Half of x is y or $\frac {1}{2}\text{x = y}$ One-third of y is z or $\frac {1}{3}\text{y = z}$
$\therefore \text{z} = \frac{1}{3}\text{y} = \frac{1}{3} (\frac{1}{2}\text{x})\times\frac{100}{100}\text{x} = \frac{16..66}{100}\text{x} = {16.66}\%\text{ of x}$

We have to just subtract $-8abc$ from $-7abc$
$= (-7abc) - (-8abc)$
$= -7abc + 8abc = abc$

A term consists of numbers and variables combined with the multiplication operation, with the variables optionally having exponents and Numerical Coefficient is often abbreviated to just coefficient. A coefficient is the numerical value in a term. If a term has no coefficient, the coefficient is an unwritten $1$ or in other words it is term without the variables.

Opening the brackets and multiplying the terms, we get $(x^2- 3x + 2) (x - 3)$
$= x^3 - 3x^2 - 3x^2 + 9x + 2x - 6$ So the term not containing $x$ is the independent term $= -6$

Since, $(9x^2 - 7) - (3x^2 + 4) = 9x^2 - 7 - ​3x^2 - 4 = 6x^2 - 11$
So, $6x^2 - 11$ should added to $3x^2 + 4$ to get $9x^2 - 7.$
Hence, the correct alternative is option $(a).$

$\Rightarrow $ Product of $x$ and $a = x \times a = ax$
$\Rightarrow $ Product of $b$ and $y = b \times y = by$
$\Rightarrow $ Product of $x$ and a subtracted from the product of $b$ and $y = by - ax$
$\therefore$ Required algebraic expression is $by - ax.$

given, $(a + b) (c + d) = ac + bc + ad + bd$ In above expression the number of terms are Four $(4)$

Degree of variables in ploynomials $(1), (3)$ and $(4)$ are not whole numbers.
$\therefore$ they are not ploynomials. While in option $(2)$ degrees of variable are whole numbers.
$\therefore$ it is a ploynomial.

$x^2 mx + 3 = (1)^2 - (2) (1) + 3 = 1 - 2 + 3 = 2$

Since, $2x^2 + 1$ has two terms $2x^2$ and $1.$
So, $2x^2 + 1$ is a binomial.
Hence, the correct alternative is option $(a).$

The given expression can be solved as shown below:
$\Rightarrow{60} = \frac{\text{b}}{4}\sqrt{{4}\times{13}^{2}}$
$\Rightarrow{60} = \frac{\text{b}}{4}\sqrt{{4}\times{169}}$
$\Rightarrow{60} = \frac{\text{b}}{4}\times\sqrt{676}$
$\Rightarrow{60} = \frac{\text{b}}{4}\times{26}$
$\Rightarrow{60} \times4 = {26}\text{ b}$
$\text{b} = \frac{240}{26} = \text{b} = 9.23$

According to classification of polynomial based on degree, a polynomial having degree $3$ is known as trinomial (cubic) polynomial.

$6x^2 - 5x^2 - 2x + 3$ has terms and $6x^3, 5x^2, 2x$ and $3,$
$\therefore$ four terms.

A polynomial having terms more than 3 is known as multinomial. for eg $-3x^4 + 2x^2+ x - 4$

The value of $(5 - 6i) - (-7i + 16) = 5 - 6i + 7i - 16 = i - 11$

Solve the polynomial as follows:$ -3(x^3 - x^2 - 2x - 5) - (4x^3 - 7x - 1)$
$= -3x^3+ 3x^2 + 6x + 15 - 4x^3 + 7x + 1$
$= -7x^3 + 3x^2 + 13x + 16$