If (x + k) is the HCF of x 2+ px + q and x 2+ qx + p, where p ≠ q, then what is the value of k ?

Question

If (x + k) is the HCF of x 2+ px + q and x 2+ qx + p, where p &ne; q, then what is the value of k ?

Prepp Team · Accepted Answer

Understanding HCF of Polynomials to Find k

The question asks us to find the value of $k$ given that $(x + k)$ is the Highest Common Factor (HCF) of two polynomials: $x^2 + px + q$ and $x^2 + qx + p$, where $p \ne q$.

The HCF of two polynomials is the polynomial of the highest degree that divides both polynomials. If $(x + k)$ is the HCF, it means $(x + k)$ is a factor of both $x^2 + px + q$ and $x^2 + qx + p$.

A fundamental property of factors and roots of a polynomial is that if $(x - a)$ is a factor of a polynomial, then $x = a$ is a root of the polynomial. This means if $(x + k)$ is a factor, then setting $x + k = 0$, we get $x = -k$ as a root of the polynomial. Therefore, $x = -k$ must be a root for both given polynomials.

Applying the Root Property to Find k

Since $x = -k$ is a root of the first polynomial, $x^2 + px + q$, substituting $x = -k$ into the polynomial must result in zero:

Let $P(x) = x^2 + px + q$.

Substitute $x = -k$:

$P(-k) = (-k)^2 + p(-k) + q$ $P(-k) = k^2 - pk + q$

Since $x = -k$ is a root, $P(-k) = 0$.

$k^2 - pk + q = 0 \quad (*).$

Similarly, since $x = -k$ is a root of the second polynomial, $x^2 + qx + p$, substituting $x = -k$ into this polynomial must also result in zero:

Let $Q(x) = x^2 + qx + p$.

Substitute $x = -k$:

$Q(-k) = (-k)^2 + q(-k) + p$ $Q(-k) = k^2 - qk + p$

Since $x = -k$ is a root, $Q(-k) = 0$.

$k^2 - qk + p = 0 \quad (**).$

Now we have two equations that both equal zero:

From $(*)$: $k^2 - pk + q = 0$

From $(**)$: $k^2 - qk + p = 0$

We can equate the expressions from $(*)$ and $(**)$ since they both equal 0:

$k^2 - pk + q = k^2 - qk + p$

Subtract $k^2$ from both sides of the equation:

$- pk + q = - qk + p$

Now, let's rearrange the terms to group terms with $k$ on one side and constant terms on the other:

Add $qk$ to both sides:

$qk - pk + q = p$

Subtract $q$ from both sides:

$qk - pk = p - q$

Factor out $k$ from the terms on the left side:

$k(q - p) = p - q$

We are given that $p \ne q$, which means $q - p \ne 0$. Therefore, we can divide both sides by $(q - p)$:

$k = \frac{p - q}{q - p}$

Notice that $p - q$ is the negative of $q - p$. That is, $p - q = -(q - p)$.

Substitute this into the expression for $k$:

$k = \frac{-(q - p)}{q - p}$

Since $q - p \ne 0$, we can cancel out the $(q - p)$ term from the numerator and the denominator:

$k = -1$

Thus, the value of $k$ is $-1$.

Summary of Steps

Recognize that if $(x+k)$ is the HCF, then $x=-k$ is a common root.
Substitute $x=-k$ into the first polynomial and set the result to 0.
Substitute $x=-k$ into the second polynomial and set the result to 0.
Equate the two resulting expressions.
Solve the equation for $k$, using the condition $p \ne q$.

Polynomial 1	Polynomial 2	Common Root
$x^2 + px + q$	$x^2 + qx + p$	$x = -k$
$P(-k) = k^2 - pk + q = 0$	$Q(-k) = k^2 - qk + p = 0$	Equate P(-k) and Q(-k)
$k^2 - pk + q = k^2 - qk + p$	$qk - pk = p - q$	$k(q - p) = p - q$
$k = \frac{p - q}{q - p}$	$k = -1$	Value of k

Revision Table: Key Concepts for HCF of Polynomials

Concept	Explanation	Relevance to Problem
Highest Common Factor (HCF)	The polynomial of the highest degree that divides two or more polynomials exactly.	$(x+k)$ is the HCF, meaning it divides both given polynomials.
Factor of a Polynomial	A polynomial $(x-a)$ is a factor of $P(x)$ if $P(a) = 0$.	If $(x+k)$ is a factor, then substituting $x=-k$ into the polynomial yields 0.
Root of a Polynomial	A value $a$ for which $P(a) = 0$. Also known as a zero of the polynomial.	If $(x+k)$ is a factor, then $x=-k$ is a root.
Solving Algebraic Equations	Manipulating equations to isolate the unknown variable using inverse operations.	Used to solve for $k$ from the equations derived by substituting the root.

Additional Information: Polynomial Roots and Factors

The relationship between roots and factors is a cornerstone of polynomial algebra. The Factor Theorem states that for a polynomial $P(x)$, $(x - a)$ is a factor if and only if $a$ is a root (or zero) of the polynomial, i.e., $P(a) = 0$. In our problem, since $(x + k)$ is the HCF, it must be a common factor, which implies $x = -k$ is a common root.

When two polynomials share a common factor, their HCF is related to that factor. If the HCF is linear like $(x+k)$, it means $x=-k$ is a root shared by both polynomials. If the HCF was quadratic, say $(x-a)(x-b)$, then $x=a$ and $x=b$ would be common roots.

The condition $p \ne q$ is crucial. If $p = q$, the two original polynomials would be identical: $x^2 + px + p$. The HCF would be the polynomial itself, $x^2 + px + p$. In this case, $(x+k)$ being the HCF would imply $x+k$ is a factor of $x^2+px+p$. If $x+k$ is a factor, then $(-k)^2 + p(-k) + p = 0$, or $k^2 - pk + p = 0$. This equation would have solutions for $k$ depending on $p$, but the HCF would not necessarily be just a linear term unless $x^2+px+p$ factors neatly into a linear term. The condition $p \ne q$ ensures the problem is well-defined and leads to a unique linear HCF like $(x+k)$ for a specific $k$.

Prepp Team · Answer

0

Prepp Team · Answer

$\frac{1}{2}$

Prepp Team · Answer

1

Polynomial 1	Polynomial 2	Common Root
\(x^2 + px + q\)	\(x^2 + qx + p\)	\(x = -k\)
\(P(-k) = k^2 - pk + q = 0\)	\(Q(-k) = k^2 - qk + p = 0\)	Equate P(-k) and Q(-k)
\(k^2 - pk + q = k^2 - qk + p\)	\(qk - pk = p - q\)	\(k(q - p) = p - q\)
\(k = \frac{p - q}{q - p}\)	\(k = -1\)	Value of k

Understanding HCF of Polynomials to Find k

Applying the Root Property to Find k

Summary of Steps

Revision Table: Key Concepts for HCF of Polynomials

Additional Information: Polynomial Roots and Factors

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Concept	Explanation	Relevance to Problem
Highest Common Factor (HCF)	The polynomial of the highest degree that divides two or more polynomials exactly.	\((x+k)\) is the HCF, meaning it divides both given polynomials.
Factor of a Polynomial	A polynomial \((x-a)\) is a factor of \(P(x)\) if \(P(a) = 0\).	If \((x+k)\) is a factor, then substituting \(x=-k\) into the polynomial yields 0.
Root of a Polynomial	A value \(a\) for which \(P(a) = 0\). Also known as a zero of the polynomial.	If \((x+k)\) is a factor, then \(x=-k\) is a root.
Solving Algebraic Equations	Manipulating equations to isolate the unknown variable using inverse operations.	Used to solve for \(k\) from the equations derived by substituting the root.