Fom 1 Lecture 2

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JEE Mathematics: Remainder & Factor Theorem (Lecture 02)

Topic 1: Remainder Theorem

  • If a polynomial $P(x)$ of degree $n \ge 1$ is divided by a linear polynomial $(x – a)$, then the remainder is $P(a)$.
  • If $P(x)$ is divided by $(ax + b)$, the remainder is $P(-b/a)$.
  • When dividing by a quadratic polynomial, the remainder will be at most a linear polynomial, i.e., $R(x) = ax + b$.
  • જો $n \ge 1$ ઘાતવાળી બહુપદી $P(x)$ ને સુરેખ બહુપદી $(x – a)$ વડે ભાગવામાં આવે, તો મળતી શેષ $P(a)$ છે.
  • જો $P(x)$ ને $(ax + b)$ વડે ભાગવામાં આવે, તો શેષ $P(-b/a)$ મળે છે.
  • જો બહુપદીને દ્વિઘાત બહુપદી વડે ભાગવામાં આવે, તો શેષ વધુમાં વધુ સુરેખ બહુપદી હોય શકે, એટલે કે $R(x) = ax + b$.
Extra Illustration

Find the remainder when the polynomial $x^3 – 2x^2 + x + 1$ is divided by $x – 2$.

જ્યારે બહુપદી $x^3 – 2x^2 + x + 1$ ને $x – 2$ વડે ભાગવામાં આવે ત્યારે શેષ શોધો.

Solution:

Let $P(x) = x^3 – 2x^2 + x + 1$.
According to the Remainder Theorem, the remainder $R$ when $P(x)$ is divided by $(x – 2)$ is $P(2)$.

$$ R = P(2) = (2)^3 – 2(2)^2 + (2) + 1 $$
$$ R = 8 – 8 + 2 + 1 = 3 $$

The remainder is $3$.

Classwork Question 1

If $P(x)=kx^{3}+3x^{2}-3$ and $Q(x)=2x^{3}-5x+k$ has same remainder when it’s divided by $(x-4)$ then find k.

બહુપદી $P(x)=kx^{3}+3x^{2}-3$ અને $Q(x)=2x^{3}-5x+k$ ને $(x-4)$ વડે ભાગતા સમાન શેષ વધે છે તો k શોધો.

Solution:

Let the polynomials be given as:
$P(x) = kx^3 + 3x^2 – 3$
$Q(x) = 2x^3 – 5x + k$

By the Remainder Theorem, the remainder when $P(x)$ and $Q(x)$ are divided by $(x – 4)$ are $P(4)$ and $Q(4)$ respectively. Since the remainders are equal:

$$ P(4) = Q(4) $$

Substitute $x = 4$ into both expressions:

$$ k(4)^3 + 3(4)^2 – 3 = 2(4)^3 – 5(4) + k $$
$$ 64k + 3(16) – 3 = 2(64) – 20 + k $$
$$ 64k + 48 – 3 = 128 – 20 + k $$
$$ 64k + 45 = 108 + k $$

Isolate $k$ on one side:

$$ 64k – k = 108 – 45 $$
$$ 63k = 63 $$
$$ k = 1 $$

Linked HW: HW Q2, HW Q4, HW Q5, HW Q6, HW Q7

Topic 2: Factor Theorem

  • A polynomial $P(x)$ has a factor $(x – a)$ if and only if $P(a) = 0$.
  • This is a special case of the Remainder Theorem where the remainder is strictly zero.
  • If two polynomials are identically equal, their corresponding coefficients must be identically equal.
  • જો અને તો જ $(x – a)$ એ બહુપદી $P(x)$ નો અવયવ છે, જો $P(a) = 0$ હોય.
  • આ શેષ પ્રમેયનો એક ખાસ કિસ્સો છે જેમાં શેષ શૂન્ય થાય છે.
  • જો બે બહુપદીઓ સમાન હોય, તો તેમના અનુરૂપ સહગુણકો પણ સમાન હોવા જોઈએ.
Extra Illustration

Find the value of $m$ if $(x – 1)$ is a factor of $x^2 + mx + 2$.

જો $(x – 1)$ એ બહુપદી $x^2 + mx + 2$ નો અવયવ હોય, તો $m$ ની કિંમત શોધો.

Solution:

Let $P(x) = x^2 + mx + 2$.
Since $(x – 1)$ is a factor of $P(x)$, by the Factor Theorem, the remainder must be zero, meaning $P(1) = 0$.

$$ P(1) = (1)^2 + m(1) + 2 = 0 $$
$$ 1 + m + 2 = 0 $$
$$ m + 3 = 0 \implies m = -3 $$

Classwork Question 2

For what values of a, $x-a$ is a factor of $x^{3}-a^{2}x+x+2$

a ની કઈ કિંમત માટે $x-a$ એ બહુપદી $x^{3}-a^{2}x+x+2$ નો અવયવ હશે.

Solution:

Let the polynomial be $P(x) = x^3 – a^2x + x + 2$.
According to the Factor Theorem, if $(x – a)$ is a factor of $P(x)$, then $P(a) = 0$.

$$ P(a) = (a)^3 – a^2(a) + (a) + 2 = 0 $$

Simplify the expression:

$$ a^3 – a^3 + a + 2 = 0 $$

The $a^3$ terms cancel out, leaving:

$$ a + 2 = 0 $$
$$ a = -2 $$

Thus, $(x – a)$ is a factor for $a = -2$.

Linked HW: HW Q1, HW Q3

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