Question

If \(-1,-2\) are two zeros of a polynomial \(2x^3+ax^2+ bx-2\). then \((a, b) =\)

• A. (1, 2)
• B. (5, 1)
• C. (3,2)
• D. (2,-1)

Answer 1

The Correct Option is B

Let the given polynomial be $P(x) = 2x^3 + ax^2 + bx - 2$. 
We are given that $-1$ and $-2$ are two zeros of the polynomial $P(x)$.

By the definition of a zero of a polynomial, if $k$ is a zero of $P(x)$, then $P(k) = 0$. 
Since $-1$ is a zero, we must have $P(-1) = 0$. 
Substituting $x = -1$ into the polynomial expression: 
$P(-1) = 2(-1)^3 + a(-1)^2 + b(-1) - 2 = 0$ 
$2(-1) + a(1) - b - 2 = 0$ 
$-2 + a - b - 2 = 0$ 
$a - b - 4 = 0$ 
$a - b = 4$ --- (Equation 1)

Since $-2$ is also a zero, we must have $P(-2) = 0$. 
Substituting $x = -2$ into the polynomial expression: 
$P(-2) = 2(-2)^3 + a(-2)^2 + b(-2) - 2 = 0$ 
$2(-8) + a(4) - 2b - 2 = 0$ 
$-16 + 4a - 2b - 2 = 0$ 
$4a - 2b - 18 = 0$ 
Divide the entire equation by 2 to simplify: 
$2a - b - 9 = 0$ 
$2a - b = 9$ --- (Equation 2)

Now we need to solve the system of two linear equations with two variables, $a$ and $b$: 
1) $a - b = 4$ 
2) $2a - b = 9$ 

To eliminate $b$, we can subtract Equation 1 from Equation 2: 
$(2a - b) - (a - b) = 9 - 4$ 
$2a - b - a + b = 5$ 
$a = 5$

Substitute the value $a = 5$ back into Equation 1: 
$a - b = 4$ 
$5 - b = 4$ 
$-b = 4 - 5$ 
$-b = -1$ 
$b = 1$

Therefore, the values of $a$ and $b$ are $a = 5$ and $b = 1$. 
The required pair $(a, b)$ is $(5, 1)$.

The correct option is (B): \((5, 1)\)