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Factor Theorem

fx=x3-6 x2+11...

Question

f(x) = x³ − 6x² + 11x − 6, g(x) = x² − 3x + 2

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Solution

It is given that and

We have

$\Rightarrow (x - 2)$ and (x − 1) are factor of g(x) by the factor theorem.

To prove that (x − 2) and (x − 1) are the factor of f(x).

It is sufficient to show that f(2) and f(1) both are equal to zero.

And

Hence, g(x) is the factor of the polynomial f(x).

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Similar questions

Use factor theorem to find whether polynomial g(x) is a factor of polynomial f(x)

f(x) = $x^{3} - 6 x^{2}$ + 11x - 6, g(x) = $x^{2}$ -3x + 2

f (x) = x^{3} - 6 x^{2} + 11 x - 6, g (x) = x + 2

Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases :

(i) f(x) = 3x + 1, x = - $\frac{1}{3}$

(ii) f(x) = $x^{2}$ - 1, x = 1, -1

(iii) g(x) = $3 x^{2}$ - 2, x = $\frac{2}{\sqrt{3}}, - \frac{2}{\sqrt{3}}$

(vi) p(x) = $x^{3} - 6 x^{2}$ + 11x - 6, x = 1,2,3

(v) f(x) = 5x - $π, x = \frac{4}{5}$

(vi) f(x) = $x^{2}$ ,x = 0

(vii) f(x) = lx + m, = - $\frac{m}{l}$

(viii) f(x) = 2x + 1, x = $\frac{1}{2}$

f (x) = 3 x + 1, x = - \frac{1}{3}

f (x) = x^{2} - 1, x = 1, - 1

g (x) = 3 x^{2} - 2, x = \frac{2}{\sqrt{3}}, - \frac{2}{\sqrt{3}}

p (x) = x^{3} - 6 x^{2} + 11 x - 6, x = 1, 2, 3

f (x) = 5 x - π, x = \frac{4}{5}

f (x) = x^{2}, x = 0

f (x) = l x + m, x = - \frac{m}{1}

f (x) = 2 x + 1, x = \frac{1}{2}

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