Q. (x^7 + 28x^4 – 42x^3 + 6x + 11).

Answer

Given expression:

\(x^7 + 28x^4 – 42x^3 + 6x + 11\)

This is a polynomial, not an equation, so there is no value of \(x\) to solve for unless the expression is set equal to \(0\).

First, check for like terms. The powers of \(x\) are different:

\(x^7,\ x^4,\ x^3,\ x,\ 1\)

So there are no like terms to combine.

Next, check for a common factor. The coefficients are:

\(1,\ 28,\ -42,\ 6,\ 11\)

The greatest common factor is \(1\), so there is no common factor to factor out.

Using the Rational Root Theorem, possible rational roots would be:

\(\pm 1,\ \pm 11\)

Test them:

\(P(1) = 1 + 28 – 42 + 6 + 11 = 4\)

\(P(-1) = -1 + 28 + 42 – 6 + 11 = 74\)

Neither equals \(0\), and \(x = 11\) and \(x = -11\) also do not give \(0\).

Final answer:

\(x^7 + 28x^4 – 42x^3 + 6x + 11\)

The polynomial is already simplified and has no simple rational linear factor.

Detailed Explanation

Problem

Analyze and factor the polynomial:

\(x^7 + 28x^4 – 42x^3 + 6x + 11\)

Step 1: Identify the expression

The expression is:

\(x^7 + 28x^4 – 42x^3 + 6x + 11\)

This is a polynomial in one variable, \(x\).

The highest power of \(x\) is \(7\), so this is a seventh-degree polynomial.

Because the problem gives an expression, not an equation, there is no single value of \(x\) to solve for unless we set the expression equal to zero.

So the natural task is to simplify or factor the polynomial as much as possible.

Step 2: Check whether there are like terms

The polynomial is:

\(x^7 + 28x^4 – 42x^3 + 6x + 11\)

The terms are:

\(x^7\)

\(28x^4\)

\(-42x^3\)

\(6x\)

\(11\)

Like terms have the same variable raised to the same power.

Here, the powers of \(x\) are different:

\(7,\ 4,\ 3,\ 1,\ 0\)

Since no two terms have the same power of \(x\), there are no like terms to combine.

So the expression is already simplified in standard polynomial form.

Step 3: Check for a common factor

A common factor must divide every term in the polynomial.

The terms are:

\(x^7,\ 28x^4,\ -42x^3,\ 6x,\ 11\)

The first four terms contain \(x\), but the constant term \(11\) does not.

So \(x\) is not a common factor of all terms.

Now check the numerical coefficients:

\(1,\ 28,\ -42,\ 6,\ 11\)

The greatest common numerical factor of these coefficients is \(1\).

Therefore, there is no common factor other than \(1\).

Step 4: Use the Rational Root Theorem

If a polynomial has a rational root, then it may have a simple linear factor.

A linear factor has the form:

\(x – r\)

where \(r\) is a root of the polynomial.

Let:

\(P(x) = x^7 + 28x^4 – 42x^3 + 6x + 11\)

The Rational Root Theorem says that possible rational roots come from:

factors of the constant term divided by factors of the leading coefficient.

The constant term is:

\(11\)

The leading coefficient is:

\(1\)

So the possible rational roots are:

\(\pm 1,\ \pm 11\)

Step 5: Test \(x = 1\)

Substitute \(x = 1\) into:

\(P(x) = x^7 + 28x^4 – 42x^3 + 6x + 11\)

We get:

\(P(1) = 1^7 + 28(1^4) – 42(1^3) + 6(1) + 11\)

Simplify each power:

\(1^7 = 1\)

\(1^4 = 1\)

\(1^3 = 1\)

So:

\(P(1) = 1 + 28 – 42 + 6 + 11\)

Add and subtract in order:

\(1 + 28 = 29\)

\(29 – 42 = -13\)

\(-13 + 6 = -7\)

\(-7 + 11 = 4\)

Therefore:

\(P(1) = 4\)

Since \(P(1) \ne 0\), \(x = 1\) is not a root.

Step 6: Test \(x = -1\)

Substitute \(x = -1\):

\(P(-1) = (-1)^7 + 28(-1)^4 – 42(-1)^3 + 6(-1) + 11\)

Simplify each power:

\((-1)^7 = -1\)

\((-1)^4 = 1\)

\((-1)^3 = -1\)

So:

\(P(-1) = -1 + 28 – 42(-1) – 6 + 11\)

Since \(-42(-1) = 42\), we get:

\(P(-1) = -1 + 28 + 42 – 6 + 11\)

Add and subtract in order:

\(-1 + 28 = 27\)

\(27 + 42 = 69\)

\(69 – 6 = 63\)

\(63 + 11 = 74\)

Therefore:

\(P(-1) = 74\)

Since \(P(-1) \ne 0\), \(x = -1\) is not a root.

Step 7: Test \(x = 11\)

Substitute \(x = 11\):

\(P(11) = 11^7 + 28(11^4) – 42(11^3) + 6(11) + 11\)

Compute the powers:

\(11^3 = 1331\)

\(11^4 = 14641\)

\(11^7 = 19487171\)

Now substitute:

\(P(11) = 19487171 + 28(14641) – 42(1331) + 66 + 11\)

Multiply:

\(28(14641) = 409948\)

\(42(1331) = 55902\)

So:

\(P(11) = 19487171 + 409948 – 55902 + 66 + 11\)

Add and subtract:

\(19487171 + 409948 = 19897119\)

\(19897119 – 55902 = 19841217\)

\(19841217 + 66 = 19841283\)

\(19841283 + 11 = 19841294\)

Therefore:

\(P(11) = 19841294\)

Since \(P(11) \ne 0\), \(x = 11\) is not a root.

Step 8: Test \(x = -11\)

Substitute \(x = -11\):

\(P(-11) = (-11)^7 + 28(-11)^4 – 42(-11)^3 + 6(-11) + 11\)

Compute the powers:

\((-11)^3 = -1331\)

\((-11)^4 = 14641\)

\((-11)^7 = -19487171\)

Now substitute:

\(P(-11) = -19487171 + 28(14641) – 42(-1331) – 66 + 11\)

Multiply:

\(28(14641) = 409948\)

\(-42(-1331) = 55902\)

So:

\(P(-11) = -19487171 + 409948 + 55902 – 66 + 11\)

Add and subtract:

\(-19487171 + 409948 = -19077223\)

\(-19077223 + 55902 = -19021321\)

\(-19021321 – 66 = -19021387\)

\(-19021387 + 11 = -19021376\)

Therefore:

\(P(-11) = -19021376\)

Since \(P(-11) \ne 0\), \(x = -11\) is not a root.

Step 9: Decide whether the polynomial has a simple rational factor

The only possible rational roots are:

\(\pm 1,\ \pm 11\)

We tested all of them:

\(P(1) = 4\)

\(P(-1) = 74\)

\(P(11) = 19841294\)

\(P(-11) = -19021376\)

None of these values equals \(0\).

Therefore, the polynomial has no rational roots.

That means it does not have a simple rational linear factor.

Step 10: Check whether the polynomial is already simplified

The polynomial is:

\(x^7 + 28x^4 – 42x^3 + 6x + 11\)

It has no like terms.

It has no common factor.

It has no rational root.

So it cannot be simplified or factored into a simple linear factor over the rational numbers.

Final answer

The polynomial is already simplified:

\(x^7 + 28x^4 – 42x^3 + 6x + 11\)

It has no common factor other than \(1\).

By the Rational Root Theorem, the only possible rational roots are:

\(\pm 1,\ \pm 11\)

After testing them, none gives \(0\).

Therefore, the polynomial has no rational roots and no simple rational linear factor.

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Algebra FAQs

What are the possible rational roots of the polynomial \(x^7 + 28x^4 - 42x^3 + 6x + 11\)?

By the Rational Root Theorem, possible rational roots are factors of 11 over factors of 1: \(\pm 1,\pm 11\). Quick substitution shows \(\pm 1\) are not roots and \(\pm 11\) give large nonzero values, so there are no rational roots.

How many positive and negative real roots can this polynomial have?

By Descartes' Rule: \(f(x)\) has at most 2 positive roots (sign changes: +,+,-,+,+) and \(f(-x)\) has at most 3 negative roots (sign changes: -,+,+,-,+). Degree 7 and odd-degree implies at least one real root in total.

Is there an easy integer or algebraic factorization for \(x^7 + 28x^4 - 42x^3 + 6x + 11\)?

There is no obvious integer linear factor (no rational roots). Nontrivial factorizations, if any, would be into higher-degree polynomials; use a CAS (e.g., sympy, Maple, Mathematica) to attempt exact factorization over integers or rationals.

How can I approximate the real roots numerically?

Use bracketing (bisection) with sign changes, Newton–Raphson for refinement, or polynomial root solvers (Durand–Kerner, Jenkins–Traub). Plotting helps find initial guesses. Software like Python+numpy, MATLAB, or WolframAlpha gives fast numerical approximations.

What is the derivative, and how does it help analyze the polynomial's shape?

The derivative is \(f'(x)=7x^6+112x^3-126x^2+6\). Solving \(f'(x)=0\) locates critical points (local extrema and inflection behavior). Numerical root-finding for \(f'\) helps determine the number and approximate locations of turning points and monotonic intervals.

What can be said about end behavior and bounds on roots?

What can be said about end behavior and bounds on roots?

How many real roots does the polynomial definitely have?

Because the degree is odd, there is at least one real root. Descartes' Rule and sign checks narrow possibilities, but exact count of real roots requires either Sturm's theorem or numerical root isolation.

What substitutions or reductions might simplify solving or factoring?

Try simple substitutions like \(x=y\) shifts won’t reduce degree. If structure suggested, test \(x^k\) substitution (e.g., for even exponents), but here exponents 7, 4, 3, 1 are not aligned. Best approach: numeric methods or CAS for algebraic factorization attempts.

Similar Questions

(x^{7} – 21x^{4} + 35x^{2} – 6x + 18). \(a^2 + b^2 = c^2\). (y^2 = x^6 + 4x^5 + 6x^4 + 2x^3 + x^2 + 2x + 1.) ( z^2 = 4 x + x^2 + 4 x^3 + 4 x^5 ). Discriminant of hyperelliptic curve \(y^2 = f(x)\). \(y^2 = x^6 + 2x^3 + 4x^2 + 4x + 1\) discriminant.
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