Is the expression ${x^{10}} + {y^3} + {t^{50}}$, polynomial in one variable or not$?$ State the reason for your answer.
Answer
${x^{10}} + {y^3} + {t^{50}}$
We can observe that in the polynomial ${x^{10}} + {y^3} + {t^{50}}$, we have $x, y$ and $t$ as the variables and the powers of $x, y$ and $t$ in each term is a whole number.
Therefore, we conclude that ${x^{10}} + {y^3} + {t^{50}}$ is a polynomial but not a polynomial in one variable.
Is the expression $y + \frac{2}{y}$, polynomial in one variable or not$?$ State the reason for your answer.
Answer
$y + \frac{2}{y}$
We can observe that in the polynomial $y + \frac{2}{y}$ ,we have $y$ as the only variable and the powers of $y$ in each term are not a whole number.
Therefore, we conclude that $y + \frac{2}{y}$ is not a polynomial in one variable.
Is the expression $3\sqrt t + t\sqrt 2$, polynomial in one variable or not? State the reason for your answer.
Answer
$3\sqrt t + t\sqrt 2$ We can observe that in the polynomial $3\sqrt t + t\sqrt 2 $ we have t as the only variable and the powers of t in each term are not a whole number. Therefore, we conclude that $3\sqrt t + t\sqrt 2$ is not a polynomial in one variable.
Is the expression ${y^2} + \sqrt 2$, polynomial in one variable or not? State the reason for your answer.
Answer
${y^2} + \sqrt 2$ We can observe that in the polynomial ${y^2} + \sqrt 2 $, we have y as the only variable and the powers of y in each term are a whole number. Therefore, we conclude that ${y^2} + \sqrt 2$ is a polynomial in one variable.
Is the expression $4{x^2} - 3x + 7{\text{ }}$, polynomial in one variable or not$?$ State the reason for your answer.
Answer
$4{x^2} - 3x + 7{\text{ }}$
We can observe that in the polynomial $4{x^2} - 3x + 7{\text{ }}$
we have $x$ as the only variable and the powers of $x$ in each term are a whole number.
Therefore, we conclude that $4{x^2} - 3x + 7{\text{ }}$ is a polynomial in one variable.
Finding a zero of $p(x),$ is the same as solving the equation $p(x) = 0$
Now, $2x + 1 = 0$ gives us $x=-\frac{1}{2}$
So, $-\frac{1}{2}$ is a zero of the polynomial $2x + 1 $
Check whether $–2$ and $2$ are zeroes of the polynomial $x + 2$
Answer
We have, $p(x) = x + 2$
Then $p(2) = 2 + 2 = 4, p(–2) = –2 + 2 = 0$
Therefore, $–2$ is a zero of the polynomial $x + 2,$ but $2$ is not the zero of the polynomial.
Using Identity $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$,
we have $(4 a-2 b-3 c)^2=[4 a+(-2 b)+(-3 c)]^2$
$=(4 a)^2+(-2 b)^2+(-3 c)^2+2(4 a)(-2 b)+2(-2 b)(-3 c)+2(-3 c)(4 a)$
$=16 a^2+4 b^2+9 c^2-16 a b+12 b c-24 a c$
This is the required expansion.
Comparing the given expression with $(x+y+z)^2$, we find that $x=3 a, y=4 b$ and $z=5 c$.
Therefore, using Identity $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$
we have $(3 a+4 b+5 c)^2=(3 a)^2+(4 b)^2+(5 c)^2+2(3 a)(4 b)+2(4 b)(5 c)+2(5 c)(3 a)$
$=9 a^2+16 b^2+25 c^2+24 a b+40 b c+30 a c$