Question

3x³ + 2x² - 19x + 6 factories using Remainder Theorem.

Class: 10 ICSC board questions. ​

Answer 1

Answer:

$\bullet \: \: \: \underline\bold{3 \: factors \: are \: (x-2)(x+3)(3x-1)}$

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Step-by-step explanation:

Given :

• 3x² + 2x² - 19x + 6

To Find :

• 3 zeros of it

Solution :

$\text{factors \: of \: 6 :} \\ ± \: 1 \\ ± \: 2 \\ ± \: 3 \\ ± \: 6$

Let us assume x = 1

$f(x) = 3x {}^{3} + 2 {x}^{2} - 19x + 6$

$\\ f(1) = 3(1) {}^{3} + 2 {(1)}^{2} - 19(1) + 6$

$\\ = 3 + 2 - 19 + 6$

$\\ = 11 - 19$

$\\ = - 8$

$\\ \therefore \: - 8 ≠0$

Let us assume x = -1

$\\ f(x) = 3 {x}^{3} + 2 {x}^{2} - 19x + 6$

$\\ f( - 1) = 3( - 1) {}^{3} + 2( - 1) {}^{2} - 19( - 1) + 6$

$\\ = - 3 + 2 + 19 + 6$

$\\ = 27 - 3$

$\\ = 24$

$\\ \therefore \: 24≠0$

Let us assume x = 2

$\\ f(x) = 3x {}^{3} + 2x {}^{2} - 19x + 6$

$\\ f(2) = 3(2) {}^{3} + 2(2) {}^{2} - 19(2) + 6$

$\\ = 24 + 8 - 38 + 6$

$\\ = 38 - 38$

$= 0$

$\\ \therefore \: 0 = 0$

Hence, x = 2

or, x - 2 = 0

$\\ \tt{x - 2) \: 3x {}^{3} + 2 {x}^{2} - 19x + 6 \: (3 {x}^{2} + 8x - 3} \\ \tt3 {x}^{3} - 6 {x}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ - - - - - - \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt 8x {}^{2} - 19x \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt 8x {}^{2} - 16x \: \: \: \: \: \: \: \: \: \: \: \: \\ - - - - - - \: \: \: \: \: \: \: \: \: \: \\ \: - \tt3x \: \: + \: \: 6 \\ \: \tt- 3x \: \: + \: \: 6 \\ \: - - - - - - \\ \: \: \: \tt0$

Quotient = 3x² + 8x - 3

$\\ \bf \: = 3 {x}^{2} + 8x - 3$

$\\ \bf = 3 {x}^{2} + 9x - x - 3$

$\\ \bf = 3x(x + 3) - 1(x + 3)$

$\\ = \bf(3x - 1)(x + 3)$

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$\\ \underline{\text{Therefore, 3 factors are (x-2)(x+3)(3x-1)}}$