Using factor theorem- show that-x-2 is a factor of x3-8

Let: fx=x3 8 Now, x 2=0 #8658;x=2 By the factor theorem, (x3 8) is a factor of the given polynomial nbsp;if nbsp;f(2) = 0. Thus, we have: f2=23 8 #160 ...

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Standard IX

Mathematics

Factor Theorem

Using factor ...

Question

Using factor theorem, show that:
(x − 2) is a factor of (x³ − 8)

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Solution

Let:
$f (x) = x^{3} - 8$
Now,
$x - 2 = 0 \Rightarrow x = 2$
By the factor theorem, (x³ $-$ 8) is a factor of the given polynomial if f(2) = 0.
Thus, we have:
$f (2) = (2^{3} - 8) = 0$
Hence, (x $-$ 2) is a factor of the given polynomial.

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Using factor theorem, show that: (x − 2) is a factor of (x3 − 8)

Let: fx=x3-8 Now, x-2=0⇒x=2 By the factor theorem, (x3 - 8) is a factor of the given polynomial if f(2) = 0. Thus, we have: f2=23-8 =0 Hence, (x - 2) is a factor of the given polynomial.

Using factor theorem, show that:
(x − 2) is a factor of (x³ − 8)

Let:
$f (x) = x^{3} - 8$
Now,
$x - 2 = 0 \Rightarrow x = 2$
By the factor theorem, (x³ $-$ 8) is a factor of the given polynomial if f(2) = 0.
Thus, we have:
$f (2) = (2^{3} - 8) = 0$
Hence, (x $-$ 2) is a factor of the given polynomial.