Polynomials in one variable

Answer: option A, k=1. The condition p(−3)=0 means that substituting x=−3 into the polynomial must give zero. Substitute carefully: p(−3)=k(−3)²+6(−3)+9. Since (−3)²=9, this becomes 9k−18+9=9k−9. Now apply the given condition: 9k−9=0, so 9k=9 and k=1. Therefore option A is correct. The most important sign point is that the square of −3 is positive, not negative. If k=2, then p(−3)=9; if k=3, it is 18; and if k=4, it is 27, so those choices do not produce zero. A polynomial value equal to zero means the substituted number is a zero or root of the polynomial, which gives a direct equation for the unknown coefficient.

If a number is a zero of a polynomial, substituting that number into the polynomial gives zero. Here x=2 is a zero, so we must use \\(p(2)=0\\). Substitution produces a simple linear equation in b, and solving it gives \\(b=-1\\). Therefore the supplied answer A is correct.

Calculate carefully: \\(p(2)=3(2)^2+b(2)-10=12+2b-10=2+2b\\). Since 2 is a zero, \\(2+2b=0\\). Hence \\(2b=-2\\) and \\(b=-1\\). Checking, \\(3(4)-2-10=12-2-10=0\\), so the value satisfies the zero condition. The important idea is to set the polynomial equal to zero, not merely substitute and stop.

The degree of a polynomial is the highest power of the variable with a non-zero coefficient. Here, the coefficient of \(x^2\) is \(a\). For the polynomial to have degree 1, this coefficient must be zero, so \(a=0\); the polynomial then becomes \(3x+5\), whose degree is 1. Exam tip: A term with a zero coefficient is omitted when determining the degree of a polynomial.

In the given polynomial, the coefficient of \\(x^2\\) is \\(-2\\), and the constant term is \\(-9\\). Therefore, their sum is \\((-2)+(-9)=-11\\). Exam tip: the constant term is the term containing no variable.

Substituting \(x=2\) in the condition \(q(2)=0\) gives \(2^2-2k+6=0\), so \(10-2k=0\). Hence, \(2k=10\) and \(k=5\), making option C correct. Exam tip: If \(a\) is a zero of a polynomial, substitute \(x=a\) and set the polynomial equal to zero.

(\frac{4}{x}=4x^{-1}) has a negative power of the variable. In a polynomial, powers of the variable must be whole numbers.

In the polynomial \(2x^3+mx^2-x+4\), the coefficient of \(x^2\) is directly \(m\). Since this coefficient is given as \(-5\), we get \(m=-5\). Exam tip: identify the coefficient attached to the required power of the variable.

The polynomial has no term containing \(x^3\), so it can be written as \(0x^3\) when all powers are considered. Therefore, the coefficient of \(x^3\) is 0. Options A and B are the coefficients of \(x^5\) and \(x^4\), while option D is the coefficient of \(x^2\). Exam tip: the coefficient of any missing power in a polynomial is 0.

A number is a zero of a polynomial if the value of the polynomial at that number is zero. \(f(2)=2^2+2(2)-8=4+4-8=0\), so 2 is a zero of the polynomial. For comparison, \(f(4)=16\), so 4 is not a zero. Exam tip: substitute each option into the polynomial and select the one that gives 0.

Substituting \(x=2\) gives \(p(2)=3(2)^2-2(2)+1=9\), while substituting \(x=1\) gives \(p(1)=3(1)^2-2(1)+1=2\). Hence, \(p(2)-p(1)=9-2=7\), so option C is correct. Option 6 results from incorrectly taking \(p(1)\) as 3. Exam tip: substitute each value carefully and simplify the squared term before subtracting.

For the polynomial to have degree 2, the coefficient of \(x^3\) must be zero so that the cubic term disappears and \(4x^2\) becomes the highest-degree term. Thus, \(a-2=0\), giving \(a=2\). For example, if \(a=4\), the coefficient of \(x^3\) is 2, so the degree remains 3. Exam tip: To reduce a polynomial’s degree, set the coefficient of its highest-power term equal to zero.

In the polynomial \(3x^2-4\), the term containing \(x\) is absent. The coefficient of an absent term is taken as \(0\), so option B is correct. In option C, the coefficient of \(x\) is \(1\), not zero. Exam tip: treat every missing term in a polynomial as having coefficient zero.

Substitute \(x=-2\) into the polynomial: \(r(-2)=(-2)^3-4(-2)=-8+8=0\). Therefore, the correct answer is 0. The distractor −4 may result from mishandling the sign in \(-4(-2)\) or in the final addition. In an exam, always use brackets when substituting a negative value.

In the polynomial \(5x^2-3x+7\), the highest power of the variable \(x\) is 2, and its leading coefficient, 5, is non-zero. Hence, its degree is 2 and it is a quadratic polynomial. It would be linear if the highest power were 1 and cubic if it were 3. Exam tip: The degree of a polynomial is the highest exponent of its variable with a non-zero coefficient.

A zero of a polynomial makes its value equal to zero, so \(p(3)=0\). Thus, \(3k+12=0\), giving \(3k=-12\) and \(k=-4\). Option A has the wrong sign; in such questions, substitute the given zero into the polynomial and equate the result to zero.

The non-zero terms are \(x^4\), \(-6x^2\), and \(9\), so the polynomial has 3 non-zero terms. Option A is incorrect because there are three terms, not two. Exam tip: Count each non-zero term separated by a plus or minus sign, including the constant term.

Substitute \(x=1\) into the polynomial: \(p(1)=1^2+a(1)+a=1+2a\). Since \(p(1)=5\), we get \(1+2a=5\), so \(2a=4\) and \(a=2\). Hence, option B is correct. Exam tip: When \(p(k)\) is given, substitute \(x=k\) directly to form an equation.

A number is a zero of a polynomial only when the value of the polynomial at that number is 0. Here, \(p(3)=3^2-9=0\) and \(p(-3)=(-3)^2-9=0\), so 3 and -3 are zeros. However, \(p(0)=0^2-9=-9\neq 0\); therefore, 0 is not a zero. Exam tip: Use \(x^2-a^2=(x-a)(x+a)\) to find the zeros quickly.

The term with the highest power of the variable is called the leading term. Here, the leading term is \(3x^4\), so its leading coefficient is 3. Although -5 is also a coefficient in the polynomial, it belongs to the lower-degree term \(-5x^2\), not the leading term. Exam tip: identify the highest power first, then select the numerical coefficient of that term.

To find \(p(3)\), substitute 3 for \(x\): \(p(3)=2(3)^2-7(3)+3=18-21+3=0\). Therefore, the correct answer is 0. Exam tip: replace \(x\) with the given value in every term; the value 18 from considering only \(2(3)^2\) is not the value of the complete polynomial.

In the polynomial \(x^3+2x^2+cx+5\), the coefficient of \(x\) is \(c\). Hence, \(c=6\), so \(c+5=6+5=11\). Option 6 is only the value of \(c\), not of \(c+5\). Exam tip: First identify the coefficient, then substitute its value in the required expression.

Substituting \(x=-2\) gives \(p(-2)=(-2)^2+4(-2)+4=4-8+4=0\). Therefore, \(-2\) is a zero of the polynomial. Option B is incorrect because \(p(0)=4\); similarly, \(p(2)=16\) and \(p(4)=36\). Exam tip: To test whether a number is a zero, substitute it into the polynomial and check whether the result is \(0\).

Substituting x=0 gives p(0)=5(0)^2-3(0)+2=2. Substituting x=1 gives p(1)=5-3+2=4. Therefore, p(0)+p(1)=2+4=6, so option C is correct. In such questions, evaluate each polynomial value separately before adding them.

In option A, x² + 3x + 1, only x appears as a variable, and its powers are non-negative integers; therefore, it is a polynomial in x and not in y. Options C and D contain both x and y, while option B is a polynomial only in y. Exam tip: identify all letters appearing as variables before deciding whether an expression is in one variable.

To find the zero of the polynomial, set (p(x)=0): (4x-12=0), which gives (4x=12) and hence (x=3). Therefore, 3 is correct. The value 12 is the constant term, not the zero. Exam tip: the zero of a linear polynomial (ax+b) is (-b/a).