Polynomials — NCERT Solutions, AI Tutor & Practice

A polynomial in one variable must have only whole-number exponents of the variable.

(i) $4x^2 - 3x + 7$: exponents are $2, 1, 0$ — all whole numbers. It is a polynomial.

(ii) $y^2 + \sqrt{2}$: exponent of $y$ is $2$; $\sqrt{2}$ is just a constant coefficient. It is a polynomial.

(iii) $3\sqrt{t} + t\sqrt{2} = 3t^{1/2} + \sqrt{2}\,t$: the term $3t^{1/2}$ has exponent $\tfrac12$, which is not a whole number. It is not a polynomial.

(iv) $x + \dfrac{2}{x} = x + 2x^{-1}$: exponent $-1$ is not a whole number. It is not a polynomial.

The coefficient of $x^2$ is the number multiplying $x^2$.

(i) $5 - x^2 + 4x$: the $x^2$ term is $-x^2 = -1\cdot x^2$, so coefficient $= -1$.

(ii) $\sqrt{2}\,x^2 - 3x$: coefficient $= \sqrt{2}$.

(iii) $\dfrac{\pi}{2}x^2 + x$: coefficient $= \dfrac{\pi}{2}$.

(iv) $\sqrt{3}\,x - 5$: there is no $x^2$ term, so coefficient $= 0$.

The degree is the highest power of the variable appearing in the polynomial.

(i) $7x^3 - 4x^2 + x - 9$: highest power is $3$, so degree $= 3$.

(ii) $5 - y^2$: highest power is $2$, so degree $= 2$.

(iii) $3$: a non-zero constant is written as $3x^0$, so degree $= 0$.

(iv) $x^4 - 2x^2 + 6$: highest power is $4$, so degree $= 4$.

A zero of $p(x)$ is the value of $x$ for which $p(x) = 0$.

Set $3x + 12 = 0$: $3x = -12 \implies x = -4.$

Verification: $p(-4) = 3(-4) + 12 = -12 + 12 = 0.$ ✓

The zero of $p(x)$ is $x = -4$.

Substitute each value into $p(x) = x^2 - 5x + 6$.

$p(0) = 0 - 0 + 6 = 6$.

$p(2) = 4 - 10 + 6 = 0$.

$p(3) = 9 - 15 + 6 = 0$.

Since $p(2) = 0$ and $p(3) = 0$, the numbers $2$ and $3$ are zeros of $p(x)$, while $0$ is not.

We split the middle term $-7x$ into two terms whose product of coefficients equals $1 \times 10 = 10$ and whose sum is $-7$.

The numbers are $-5$ and $-2$ (since $(-5)(-2)=10$, $-5+(-2)=-7$).

$x^2 - 7x + 10 = x^2 - 5x - 2x + 10 = x(x - 5) - 2(x - 5) = (x - 5)(x - 2).$

Setting each factor to zero: $x - 5 = 0 \Rightarrow x = 5$ and $x - 2 = 0 \Rightarrow x = 2$.

The zeros are $x = 5$ and $x = 2$.

By the Remainder Theorem, the remainder when $p(x)$ is divided by $(x - a)$ is $p(a)$.

Here $a = 2$, so compute $p(2)$: $p(2) = (2)^3 - 3(2)^2 + 4(2) - 5 = 8 - 12 + 8 - 5 = -1.$

The remainder is $-1$.

First find the value of $x$ that makes the divisor zero: $2x - 1 = 0 \implies x = \tfrac12.$

By the Remainder Theorem, remainder $= p\!\left(\tfrac12\right)$: $p\!\left(\tfrac12\right) = 4\left(\tfrac18\right) - 12\left(\tfrac14\right) + 14\left(\tfrac12\right) - 3.$ $= \tfrac12 - 3 + 7 - 3 = \tfrac12 + 1 = \tfrac32.$

The remainder is $\dfrac{3}{2}$.

By the Factor Theorem, $(x + 1) = (x - (-1))$ is a factor of $p(x)$ if and only if $p(-1) = 0$.

$p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0.$

Since $p(-1) = 0$, $(x + 1)$ is a factor of $p(x)$.

If $(x - 2)$ is a factor, then by the Factor Theorem $p(2) = 0$.

$p(2) = 2(2)^3 - 3(2)^2 + k(2) - 2 = 16 - 12 + 2k - 2 = 2 + 2k.$

Set equal to zero: $2 + 2k = 0 \implies 2k = -2 \implies k = -1.$

So $k = -1$.

Use the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$ with $a = 2x$ and $b = 3y$.

$a^3 = (2x)^3 = 8x^3, \qquad b^3 = (3y)^3 = 27y^3.$ $3ab(a+b) = 3(2x)(3y)(2x + 3y) = 18xy(2x + 3y) = 36x^2 y + 54 x y^2.$

Therefore: $(2x + 3y)^3 = 8x^3 + 27y^3 + 36x^2 y + 54 x y^2.$

Rewrite the expression to match $a^3 + b^3 + c^3 - 3abc$.

Let $a = x$, $b = -2y$, $c = 3z$. Then: $a^3 = x^3,\quad b^3 = (-2y)^3 = -8y^3,\quad c^3 = (3z)^3 = 27z^3.$ $-3abc = -3(x)(-2y)(3z) = 18xyz.$

So the expression is exactly $a^3 + b^3 + c^3 - 3abc$.

Using the identity $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$:

$= (x - 2y + 3z)\big(x^2 + 4y^2 + 9z^2 + 2xy + 6yz - 3zx\big).$

Let $a = -12$, $b = 7$, $c = 5$. Notice: $a + b + c = -12 + 7 + 5 = 0.$

When $a + b + c = 0$, the identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(\ldots)$ gives $a^3 + b^3 + c^3 = 3abc.$

Therefore: $(-12)^3 + 7^3 + 5^3 = 3(-12)(7)(5) = 3 \times (-420) = -1260.$

Let $p(x) = x^3 - 6x^2 + 11x - 6$. Try small factors of $6$.

$p(1) = 1 - 6 + 11 - 6 = 0$, so $(x - 1)$ is a factor.

Divide $p(x)$ by $(x - 1)$ to get the quotient: $x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6).$

Now factorise the quadratic by splitting the middle term: $x^2 - 5x + 6 = x^2 - 2x - 3x + 6 = (x - 2)(x - 3).$

Therefore: $x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3).$