Use the factor theorem and state whether $g (x)$ is a factor of $p (x)$ in the following case. State True Or False:
$p (x) = x^{3} - 4 x^{2} + x + 6; g (x) = x - 3$

True

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False

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Solution

The correct option is A True
The zero of $g (x) = x - 3$ is $3$
Here, $p (x) = x^{3} - 4 x^{2} + x + 6$
Now, $p (3) = (3)^{3} - 4 (3)^{2} + 3 + 6$
$= 27 - 36 + 3 + 6 = 0$
Since, $p (3) = 0$
Therefore, by Factor theorem, $g (x)$ is a factor of $p (x)$
Hence the answer is True

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Similar questions

Use the factor theorem to determine and state whether $g (x)$ is a factor of $p (x)$ in each of the following cases:

p (x) = x^{3} + 3 x^{2} + 3 x + 1, g (x) = x + 2

Q. Question 2
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p (x) = 2 x^{3} + x^{2} - 2 x - 1, g (x) = x + 1

(i i) p (x) = x^{3} + 3 x^{2} + 3 x + 1, g (x) = x + 2

(i i i) p (x) = x^{3} - 4 x^{2} + x + 6, g (x) = x - 3

Q. In the following case, use factor theorem to find whether

g (x)

is a factor of the polynomial

p (x)

or not.

p (x) = 2 x^{4} + x^{3} + 4 x^{2} - x - 7 g (x) = x + 2

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
$(i) p (x) = 2 x^{3} + x^{2} - 2 x - 1, g (x) = x + 1$
$(i i) p (x) = x^{3} + 3 x^{2} + 3 x + 1, g (x) = x + 2$ [4 MARKS]

Q. Use the Factor Theorem to determine whether

g (x)

is factor of

f (x)

in each of the following cases:

f (x) = x^{3} - 4 x^{2} + x + 6, g (x) = x - 2

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Use the factor theorem and state whether g(x) is a factor of p(x) in the following case. State True Or False:p(x)=x3−4x2+x+6; g(x)=x−3

The correct option is A TrueThe zero of g(x)=x−3 is 3Here, p(x)=x3−4x2+x+6Now, p(3)=(3)3−4(3)2+3+6 =27−36+3+6=0Since, p(3)=0Therefore, by Factor theorem, g(x) is a factor of p(x)Hence the answer is True

Use the factor theorem and state whether $g (x)$ is a factor of $p (x)$ in the following case. State True Or False:
$p (x) = x^{3} - 4 x^{2} + x + 6; g (x) = x - 3$

The correct option is A True
The zero of $g (x) = x - 3$ is $3$
Here, $p (x) = x^{3} - 4 x^{2} + x + 6$
Now, $p (3) = (3)^{3} - 4 (3)^{2} + 3 + 6$
$= 27 - 36 + 3 + 6 = 0$
Since, $p (3) = 0$
Therefore, by Factor theorem, $g (x)$ is a factor of $p (x)$
Hence the answer is True