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Q2(iii):
Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: (iii) $p(x) = x^3 – 4x^2 + x + 6, g(x) = x – 3$

Solution :

Given Variables & Theoretical Foundation

We are given the following polynomial and a linear divisor:

  • Dividend Polynomial: $p(x) = x^3 - 4x^2 + x + 6$
  • Divisor Polynomial: $g(x) = x - 3$

Theoretical Justification [The Factor Theorem]: The Factor Theorem states that a polynomial $p(x)$ has a factor $(x - c)$ if and only if $p(c) = 0$. Therefore, to determine if $g(x)$ is a factor of $p(x)$, we must evaluate the polynomial at the root of $g(x)$ and check if the remainder is zero.

Step 1: Determining the Root of the Divisor

To find the value of $c$ to substitute into $p(x)$, we set the divisor $g(x)$ equal to zero and solve for $x$:

$g(x) = 0$
$x - 3 = 0$
$x = 3$

Here, our test value is $c = 3$.

Step 2: Evaluating the Polynomial $p(x)$ at $x = 3$

We substitute $x = 3$ into the original polynomial $p(x)$ to find the remainder [Per the Remainder Theorem]:

$p(3) = (3)^3 - 4(3)^2 + (3) + 6$

Now, we perform the arithmetic operations step-by-step:

  • Calculate the cube of 3: $(3)^3 = 27$
  • Calculate the square of 3, then multiply by -4: $-4(3)^2 = -4(9) = -36$
  • Substitute these values back into the expression:

$p(3) = 27 - 36 + 3 + 6$

Group the positive and negative terms to simplify the addition:

$p(3) = (27 + 3 + 6) - 36$
$p(3) = 36 - 36$
$p(3) = 0$

Step 3: Graphical Verification of the Roots

Because $p(3) = 0$, $x = 3$ is an $x$-intercept (or root) of the polynomial curve $y = x^3 - 4x^2 + x + 6$. The graph below plots the exact coordinates of the polynomial, visually confirming that the curve intersects the $x$-axis exactly at $x = 3$.

x p(x) x = -1 x = 2 x = 3 y-int (0, 6)

Step 4: Logical Conclusion

Since the evaluation of the polynomial at $x = 3$ yields a remainder of exactly $0$ ($p(3) = 0$), the condition of the Factor Theorem is perfectly satisfied.

Final Solution: By the Factor Theorem, since $p(3) = 0$, $g(x) = x - 3$ is indeed a factor of the polynomial $p(x) = x^3 - 4x^2 + x + 6$.


More Questions from Class 9 Mathematics Polynomials EXERCISE 2.3


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