If 1 and 3 are the zeros of the polynomial p( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that $1$ and $3$ are zeros of $p(x)$,$(x - 1)$ and $(x - 3)$ are factors of $p(x)$.Therefore,$(x - 1)(x - 3) = x^2 - 4x + 3$ is a factor of

Vedclass · Accepted Answer

$-3, \frac{5}{2}$

Vedclass · Answer

(C) Given that $1$ and $3$ are zeros of $p(x)$,$(x - 1)$ and $(x - 3)$ are factors of $p(x)$.Therefore,$(x - 1)(x - 3) = x^2 - 4x + 3$ is a factor of $p(x)$.Dividing $p(x) = 2x^4 - 7x^3 - 13x^2 + 63x - 45$ by $(x^2 - 4x + 3)$:$2x^4 - 7x^3 - 13x^2 + 63x - 45 = (x^2 - 4x + 3)(2x^2 + x - 15)$.Now,factorize the quadratic $2x^2 + x - 15$:$2x^2 + 6x - 5x - 15 = 2x(x + 3) - 5(x + 3) = (2x - 5)(x + 3)$.Setting these factors to zero,we get $x = -3$ and $x = \frac{5}{2}$.Thus,the remaining zeros are $-3$ and $\frac{5}{2}$.

If $1$ and $3$ are the zeros of the polynomial $p(x) = 2x^4 - 7x^3 - 13x^2 + 63x - 45$,then find the remaining zeros of $p(x)$.

Similar Questions

If two zeros of the polynomial $p(x) = x^3 + x^2 - 5x - 5$ are $\sqrt{5}$ and $-\sqrt{5}$,then the third zero is . . . . . . .

If $\alpha, \beta$,and $\gamma$ are the zeros of the cubic polynomial $p(x) = ax^3 + bx^2 + cx + d$ where $a \neq 0$,then the product of its zeros $(\alpha \beta \gamma)$ is:

If the zero of the polynomial $p(x) = 3x^{3} - x^{2} - ax - 45$ is $3$,then $a = \dots$

The value of $p(x) = 4x^3 + 3x^2 + 2x + 1$ at $x = 1$ is $\ldots \ldots \ldots$

Examine the validity of the statement: $(x+3)$ is a factor of $x^{3}+6x^{2}+5x-12$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module