Polynomials in one variable

For a monic quadratic with zeroes α and β, the required polynomial is x² − (α + β)x + αβ. Let α = 1 + √2 and β = 1 − √2. Their sum is α + β = 2 because the irrational terms cancel. Their product is αβ = (1 + √2)(1 − √2) = 1 − 2 = −1, using the difference-of-squares identity. Hence the polynomial is x² − 2x − 1, so option A is correct and its coefficients are rational. Option B has the wrong sign for the linear coefficient. Option C has neither the correct sum-product pair nor the correct constant term. Option D has constant term 3, whereas the product of the zeroes must be −1.

For the degree to be (3), the coefficient of (x^5) must be (0), so (a-5=0). First make the highest-degree term vanish.

Substituting \(x=-1\) gives \(p(-1)=k(-1)^2-5(-1)+6=k+5+6=k+11\). Thus, \(k+11=14\), so \(k=3\). For instance, taking \(k=2\) gives 13, not 14. In the exam, check both \((-1)^2\) and the sign in \(-5(-1)\) carefully.

When adding polynomials, coefficients of like powers are added. The coefficients of x² are −2 and 7, so the coefficient in the sum is −2 + 7 = 5. Option C gives only the x² coefficient of q(x), not of the complete sum. Exam tip: group terms having the same power before adding polynomials.

To find the degree of a sum, first add like terms and then inspect the highest power whose coefficient is not zero. Adding the polynomials gives (4x^4-3x^2+2)+(-4x^4+5x^3+x-8). The x^4 terms cancel because 4x^4-4x^4=0. The remaining polynomial is 5x^3-3x^2+x-6.

The highest power still present is x^3, whose coefficient is 5, not zero. Therefore the degree of the sum is 3, so choice C is correct. A common error is to retain degree 4 merely because both original polynomials had degree 4. Degree can decrease after addition when leading terms cancel, so the resulting expression must always be simplified before its degree is stated.

The direct answer is option B, 7. The degree of a non-zero polynomial is the highest power with a non-zero coefficient. If the leading terms of p and q have degrees 4 and 3, they look like \\(ax^4\\) and \\(bx^3\\), where a and b are non-zero. Their product has leading term \\(abx^{4+3}=abx^7\\). Since the product of two non-zero numbers a and b is non-zero, this seventh-degree term cannot disappear. Therefore \\(\\deg(pq)=4+3=7\\), so option B is correct. Option A, 1, is not the degree rule for multiplication. Option C, 12, comes from multiplying 4 and 3, but degrees are added, not multiplied. Option D, 4, is only the degree of p and ignores q. This rule assumes both polynomials are non-zero, as stated generally here. Memory cue: multiplication of polynomials means addition of their degrees.

((x-2)(3x^2+4x-5)=3x^3-2x^2-13x+10), so the coefficient of (x^2) is (-2). Combine like terms after expansion.

Since the polynomial is zero at its zero, \(p(3)=0\). Thus, \(3^2-3k+12=0\), giving \(21-3k=0\) and hence \(k=7\). Therefore, 7 is correct. Exam tip: whenever a zero is given, substitute it in the polynomial and set the result equal to zero.

Substitute x = 2 into the polynomial: p(2) = 2(2³) − 9(2²) + 12(2) − 5 = 16 − 36 + 24 − 5 = −1. Hence, option A is correct. Option B may result from an error in the final addition and subtraction. Exam tip: after substitution, evaluate powers first, then multiplication, and finally addition or subtraction.

Substituting \(x=1\) gives \(p(1)=1^3-4(1)^2+m(1)+6=1-4+m+6=m+3\). Since \(p(1)=0\), we require \(m+3=0\), so \(m=-3\). The nearby option \(-2\) would give \(p(1)=1\), not zero. Exam tip: when a value is given as a zero of a polynomial, substitute it and set the resulting expression equal to zero.

Substituting -1 for x gives p(-1) = (-1)^4 - 2(-1)^2 + 1 = 1 - 2 + 1 = 0. Since even powers of -1 equal 1, treating x^2 as -1 would incorrectly give -2. Exam tip: Always enclose a negative value in brackets before raising it to a power.

In a polynomial, every exponent of x must be a non-negative integer. In option A, the highest exponent is 5, while \(\sqrt{2}\) is a valid real coefficient. Option B contains \(1/x=x^{-1}\), so it is not a polynomial. Exam tip: check exponents before checking coefficients.

The direct answer is C: \(x^2+\frac{1}{x}+4\) is not a polynomial in \(x\). A polynomial in one variable has only non-negative whole-number powers of the variable, such as \(x^0,x^1,x^2\), and its variable must not occur in a denominator. Here \(\frac{1}{x}=x^{-1}\), whose exponent is negative, so the expression fails the definition. Option A, \(x^2+2x+1\), is a polynomial because its powers are 2, 1 and 0. Option B, \(2x^2-5x+3\), is also a polynomial for the same reason. Option C is correct because of the term \(1/x\). Option D, \(7x^2-9\), has powers 2 and 0 and is therefore a polynomial. Remember: a variable under a denominator or inside a non-whole power is an immediate warning sign.

The sum of all coefficients of a polynomial is found by evaluating it at \(x=1\). Here, \(p(1)=5(1)^3-2(1)^2+1-4=5-2+1-4=0\). Therefore, the correct answer is 0. Exam tip: use \(p(1)\) for the sum of coefficients, not \(p(0)\), which gives only the constant term.

The powers in \(3x^4\), \(-5x^2\), and \(-7=-7x^0\) are 4, 2, and 0, respectively, and all are even. Only \(2x=2x^1\) has an odd power, so the required sum is 2. Exam tip: the constant term has power 0, which is even, not odd.

To find the constant term, multiply the constant terms of the two factors. The constant terms are -3 in (2x-3) and 1 in (x^2+x+1), so the constant term is (-3)·1 = -3. Option C, 1, is only the constant term of the second factor, not of the product. Exam tip: In a product of polynomials, the constant term is obtained by multiplying the constant terms of all factors.

\(x^2(x^2+1)=x^4+x^2\), so the highest power of the variable is 4 and all powers are non-negative integers. Option C has degree 5, while B and D involve fractional and negative powers. Exam tip: check exponents first.

Using \(f(1)=4\), we get \(1+a+b=4\), so \(a+b=3\). Using \(f(-1)=-2\), we get \(-1-a+b=-2\), so \(-a+b=-1\). Subtracting the second equation from the first gives \(2a=4\), hence \(a=2\). Therefore, option B is correct. Exam tip: when values at \(1\) and \(-1\) are given, substitute them directly and eliminate the common constant term.

Substituting \(x=0\) gives \(p(0)=s=5\). Then, substituting \(x=1\), \(p(1)=1+r+5=9\), so \(r=3\). Therefore, option C is correct. Exam tip: when evaluating a polynomial at zero, all terms containing \(x\) vanish, leaving only the constant term.

For a constant polynomial, both (m+1=0) and (m-2=0) are needed, which is impossible together. All variable terms must vanish.

Substituting the values gives \(p(3)=3^2+2(3)-8=7\) and \(p(-2)=(-2)^2+2(-2)-8=-8\). Therefore, \(p(3)-p(-2)=7-(-8)=15\). Exam tip: when subtracting \(p(-2)\), subtract the negative value correctly, which becomes addition.

To find \(p(2)\), substitute 2 for \(x\): \(p(2)=2^3-8=8-8=0\). Hence, 2 is a zero of the polynomial. Option B incorrectly uses \(3^2\) instead of \(2^3\), while option C changes the exponent from 3 to 2. In an exam, remember that a number is a zero of a polynomial exactly when substitution gives a value of 0.

Substituting -1 for x gives p(-1)=(-1)^3+(-1)^2+(-1)+1=-1+1-1+1=0. Therefore, option B is correct. Exam tip: always use parentheses when substituting a negative number, because ignoring them can cause a sign error and lead to an answer such as -2.

Since \(p(0)=c=5\), substituting \(x=1\) gives \(p(1)=2+b+5=1\). Hence, \(b+7=1\) and \(b=-6\). The distractor \(-5\) results from an arithmetic error while solving this equation. Exam tip: for a polynomial, the constant term is obtained directly by evaluating it at \(x=0\).

The polynomial \(p(x)=x^2-5x+6=(x-2)(x-3)\). Therefore, \(p(2)=0\) and \(p(3)=0\), since 2 and 3 are its zeroes. Hence, \(p(2)+p(3)=0+0=0\). Exam tip: If a number is a zero of a polynomial, the value of the polynomial at that number is directly 0.