If x^2-5x+6 is a factor of f(x)=x^4-17x^3+kx^ | vedclass

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

Question

(B) Given that $x^2-5x+6$ is a factor of $f(x)=x^4-17x^3+kx^2-247x+210$. We can factorize $x^2-5x+6$ as $(x-2)(x-3)$. Since $(x-2)$ and $(x-3)$ are fa

Vedclass · Accepted Answer

$x^2-12x+35$

Vedclass · Answer

(B) Given that $x^2-5x+6$ is a factor of $f(x)=x^4-17x^3+kx^2-247x+210$. We can factorize $x^2-5x+6$ as $(x-2)(x-3)$. Since $(x-2)$ and $(x-3)$ are factors of $f(x)$,we have $f(2)=0$ and $f(3)=0$. Let the other quadratic factor be $x^2+ax+b$. Since the leading coefficient of $f(x)$ is $1$ and the constant term is $210$,we have $(x^2-5x+6)(x^2+ax+35) = x^4-17x^3+kx^2-247x+210$. Comparing the coefficient of $x^3$: $a-5 = -17$,which gives $a = -12$. Thus,the other quadratic factor is $x^2-12x+35$.

If $x^2-5x+6$ is a factor of $f(x)=x^4-17x^3+kx^2-247x+210$,then the other quadratic factor of $f(x)$ is

Similar Questions

If $(1 - p)$ is a root of the quadratic equation $x^2 + px + (1 - p) = 0$,then its roots are:

Solve the equation $-x^{2}+x-2=0$.

Let $a, b, c$ be three real numbers such that $a + 2b + 4c = 0$. Then the equation $ax^2 + bx + c = 0$

If the roots of the equations $ax^2 + 2bx + c = 0$ and $bx^2 - 2\sqrt{ac}x + b = 0$ are real,then:

The roots of the equation $\sqrt{3x + 1} + 1 = \sqrt{x}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module