The HCF of 6 x 2-4 and 15 x -2 x -7 is-A- 3 x -2B- 3 x -2 x -7C- 5 x -2D- 3 x -2 x -7

The correct option is A 3(x 2)6(x^2 4) = 2× 3× (x 2)× (x+2). 15(x+2)(x+7) = 3× 5× (x+2)× (x+7). Hence, the common irreducible factors between 6(x^2 4) and 15 ...

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Standard X

Mathematics

GCD of Polynomials

The HCF of 6 ...

Question

$The HCF of 6 (x^{2} - 4) and 15 (x + 2) (x + 7) is :$

3 (x - 2)

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3 (x - 2) (x + 7)

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3 (x + 2) (x + 7)

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5 (x - 2)

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Solution

The correct option is A $3 (x - 2)$
$6 (x^{2} - 4) = 2 \times 3 \times (x - 2) \times (x + 2)$ .
$15 (x + 2) (x + 7) = 3 \times 5 \times (x + 2) \times (x + 7)$ .
Hence, the common irreducible factors between $6 (x^{2} - 4) a n d 15 (x + 2) (x + 7)$ are 3 and (x-2).
The least exponents of these factors are respectively 1 and 1.
Hence, the HCF = $3^{1} \times (x - 2)^{1} = 3 (x - 2)$ .

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The HCF of 6(x2−4) and 15(x+2)(x+7) is:

The correct option is A 3(x−2)6(x2−4)=2× 3× (x−2)× (x+2). 15(x+2)(x+7)=3× 5× (x+2)× (x+7). Hence, the common irreducible factors between 6(x2−4) and 15(x+2)(x+7) are 3 and (x-2). The least exponents of these factors are respectively 1 and 1. Hence, the HCF = 31× (x−2)1=3(x−2).

$The HCF of 6 (x^{2} - 4) and 15 (x + 2) (x + 7) is :$