Polynomials in one variable

In the term
kx
, the coefficient of
x
is
k
. Therefore, if the coefficient of
x
is 5, then
k=5
. The value 8 is the constant term, not the coefficient of
x
. Exam tip: identify the factor multiplying the specified variable to find its coefficient.

To find \(p(3)\), substitute 3 for \(x\): \(p(3)=3^2-9=9-9=0\). Hence, the correct answer is 0. The value 9 is only \(3^2\); the subtraction of 9 must also be performed. In exams, directly substitute the given value of the variable into the polynomial.

A zero of a polynomial is a value that makes the polynomial equal to zero. For option A, \(p(2)=2-2=0\), so \(x=2\) is its zero. In contrast, option B gives \(p(2)=2+2=4\), so it is not correct. Exam tip: Substitute the given value into each polynomial and check whether the result is zero.

Substitute \(x=2\) into the polynomial: \(p(2)=2^2+2(2)-8=4+4-8=0\). Therefore, the correct answer is 0. Since the polynomial’s value is zero at \(x=2\), 2 is also a zero of the polynomial. Exam tip: substitute the given value carefully before simplifying.

The degree of a polynomial is the highest exponent of the variable whose coefficient is non-zero. In \(ax+b\), the condition \(a\ne0\) ensures that the term \(ax\) is present, so the highest power of \(x\) is 1. Therefore, its degree is 1. Option 0 applies to a non-zero constant polynomial, not a linear polynomial. Exam tip: The general form of a linear polynomial is \(ax+b\) with \(a\ne0\).

The term with the highest power in the polynomial is \(9y^4\). Its coefficient is 9, so the leading coefficient is 9. The value -1 in option B is the coefficient of the \(y^2\) term, not the leading term. Exam tip: identify the highest-degree term first, then take its coefficient.

The term containing \\(x^3\\) is \\(4x^3\\), so its coefficient is 4. The terms \\(-2x^2\\), \\(7x\\), and \\(-1\\) belong to different powers or are the constant term. Exam tip: identify the number multiplying the required power of the variable.

A polynomial is in standard form when its terms are arranged from the greatest exponent of the variable to the least, usually placing the constant term last. In option B, the powers are 3, 1, and 0, which decrease from left to right: 5x³ + 2x - 1. Option A is not descending, C is ascending overall, and D uses increasing powers.

When adding polynomials, combine terms with the same power of the variable: \(3x^2+x^2=4x^2\), \(-4x+2x=-2x\), and \(1-5=-4\). Therefore, the sum is \(4x^2-2x-4\). Option B results from incorrectly combining coefficients, while option D incorrectly multiplies powers instead of adding the polynomials. Exam tip: arrange the \(x^2\), \(x\), and constant terms in separate groups before adding.

We need to find \(p(x)-r(x)\): \((5x^2+3x-2)-(2x^2-x+4)\). The minus sign changes the sign of every term in the second polynomial, giving \(5x^2-2x^2=3x^2\), \(3x-(-x)=4x\), and \(-2-4=-6\). Therefore, the result is \(3x^2+4x-6\). Exam tip: when subtracting a polynomial, distribute the minus sign to every term inside the second bracket before combining like terms.

Using the distributive property, multiply \(2x\) by every term in the bracket: \(2x\cdot3x^2=6x^3\), \(2x\cdot(-5x)=-10x^2\), and \(2x\cdot1=2x\). Hence, the simplified form is \(6x^3-10x^2+2x\). In option B, the factor \(x\) has not been multiplied correctly into each term. Exam tip: distribute the outside factor to every term and carefully preserve the signs.

Using the distributive property,
(x+3)(x-4)=x^2-4x+3x-12=x^2-x-12. Therefore, option A is correct. In option B, the middle terms have been combined incorrectly. Exam tip: multiply every term in one binomial by both terms in the other binomial, then combine like terms.

Using the distributive property, multiply each term by every term: (2x−1)(x+5) = 2x² + 10x − x − 5 = 2x² + 9x − 5. Therefore, option A is correct. Option B does not combine the −x and −5 terms correctly, while option D has the wrong sign for the constant term. Exam tip: In binomial multiplication, write all four products before combining like terms.

To find \(p(1)\), substitute \(x=1\) in the polynomial: \(p(1)=1^3-6(1)^2+11(1)-6=1-6+11-6=0\). Hence, 0 is the correct value, and 1 is also a zero of the polynomial. Exam tip: when evaluating a polynomial, replace every occurrence of \(x\) with the given value, including powers.

Since \(p(2)=0\), substitute \(x=2\) into the polynomial: \(2^2+2k+6=0\), so \(4+2k+6=0\). Hence \(2k=-10\) and \(k=-5\). Therefore, option A is correct. Exam tip: whenever a zero of a polynomial is given, substitute it for \(x\) and set the polynomial equal to zero.

Substituting \(x=5\) in the given polynomial gives \(p(5)=25-5a+10=0\). Thus, \(35-5a=0\), so \(5a=35\) and \(a=7\). If \(a=10\), then \(p(5)=25-50+10=-15\), so that option is incorrect. Exam tip: substitute the given zero directly into the polynomial.

If \(x=1\) is a zero of the polynomial, its value at \(x=1\) must be zero. Thus, \(1^2-4(1)+m=0\), which gives \(1-4+m=0\) and hence \(m=3\). Therefore, option C is correct. Exam tip: Substitute the given zero into the polynomial and equate the result to zero.

Substituting x = 1 gives p(1) = 2(1)^2 - 3(1) + 1 = 2 - 3 + 1 = 0. A value of x for which a polynomial becomes zero is called a zero of that polynomial; therefore, 1 is a zero. Option B incorrectly evaluates p(1), while the degree and leading coefficient of the polynomial are both 2, not 1. Exam tip: To test whether a number is a zero, substitute it in the polynomial and check whether the result is 0.

Substituting (x=-2) gives (p(-2)=(-2)^2+4(-2)+4=4-8+4=0), so option A is correct. Choosing 2 or 4 usually results from mishandling the signs after substitution or addition. In an exam, always use parentheses when squaring a negative value.

Option B, \(4x^3+x-9\), has only non-negative integer powers of the variable \(x\), so it is a polynomial. Its highest power is 3, making it a cubic polynomial rather than a quadratic polynomial. Option A has degree 2 and is therefore quadratic. Option C contains \(x^{-1}\), and option D contains \(x^{1/2}\); hence neither is a polynomial. Exam tip: check that all variable exponents are non-negative integers, then identify the highest exponent as the degree.