Question

I need step by step solution of this question.i will give brainliest to who gives step by step answer.​

Answer 1

Answer:

• (- 1, - 2/3, 1/2)

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Factorize as below and find roots:

• 6x³ + 7x² - x - 2 = 0
• 6x³ + 6x² + x² - x - 2 = 0
• 6x²(x + 1) + x² + x - 2x - 2 = 0
• 6x²(x + 1) + x(x + 1) - 2(x + 1) = 0
• (x + 1)(6x² + x - 2) = 0
• (x + 1)(6x² + 4x - 3x - 2) = 0
• (x + 1)(2x(3x + 2) - (3x + 2)) = 0
• (x + 1)(3x + 2)(2x - 1) = 0

• x + 1 = 0 or 3x + 2 = 0 or 2x - 1 = 0
• x = - 1 or x = - 2/3 or x = 1/2

Answer 2

Answer:

$x=-1, \quad x=\dfrac{1}{2}, \quad x=-\dfrac{2}{3}$

Step-by-step explanation:

Given polynomial:

$f(x)=6x^2+7x^2-x-2$

$\boxed{\begin{minipage}{6 cm}\underline{Factor Theorem}\\\\If f(x) is a polynomial, and f(a) = 0 ,\\ then (x-a) is a factor of f(x) .\\\end{minipage}}$

Since the constant of the given polynomial is -2, the possible factors are ±1 and ±2.

Apply the Factor Theorem to check the possible values:

$f(1)=6(1)^3+7(1)^2-(1)-2=10 \neq 0$

$f(-1)=6(-1)^3+7(-1)^2-(-1)-2=0$

$f(2)=6(2)^3+7(2)^2-(2)-2=72 \neq 0$

$f(-2)=6(-2)^3+7(-2)^2-(-2)-2=-20 \neq 0$

Therefore, (x + 1) is a factor of the polynomial:

$f(x)=(x+1)(ax^2+bx+c)$

Compare with the original function to find the coefficients of the quadratic:

$6x^3=ax^3 \implies a=6$

$7x^2=(b+a)x^2 \implies b=1$

$-2=c \implies c=-2$

Therefore:

$f(x)=(x+1)(6x^2+x-2)$

Factor the quadratic:

$f(x)=(x+1)(6x^2+4x-3x-2)$

$f(x)=(x+1)[2x(3x+2)-1(3x+2)]$

$f(x)=(x+1)(2x-1)(3x+2)$

Set the polynomial function to zero:

$(x+1)(2x-1)(3x+2)=0$

Apply the zero product property:

$(x+1)=0 \implies x=-1$

$(2x-1)=0 \implies x=\dfrac{1}{2}$

$(3x+2)=0 \implies x=-\dfrac{2}{3}$

Therefore, the solutions are:

$x=-1, \quad x=\dfrac{1}{2}, \quad x=-\dfrac{2}{3}$