Polynomials in one variable

Substituting \(x=-1\), \(p(-1)=(-1)^3-4(-1)^2+(-1)+6=-1-4-1+6=0\). Therefore, 0 is the correct value, and \(-1\) is a zero of the polynomial. Choosing 2 usually results from mishandling the sign or the square of \(-1\). In an exam, always place a negative substitution inside parentheses.

Since \(p(3)=0\), substitute \(x=3\) into the polynomial: \(p(3)=3^2+3k-15=0\). Thus, \(9+3k-15=0\), so \(3k=6\) and \(k=2\). The value 3 does not make the polynomial equal to zero. Exam tip: directly substitute the given zero of the polynomial for \(x\).

Substituting \(x=3\) in the condition \(p(3)=0\) gives \(3^2-3a+18=0\), or \(27-3a=0\). Hence, \(3a=27\) and \(a=9\). Exam tip: when the value of a polynomial is given, substitute the specified value of \(x\) directly and solve the resulting equation.

If \(2\) is a zero of the polynomial, substituting \(x=2\) must make its value zero: \(2^2-6(2)+m=0\), so \(4-12+m=0\) and \(m=8\). Exam tip: Substitute the given zero into the polynomial and equate the result to zero.

p(1)=3(1)²−10(1)+7=3−10+7=0. Therefore, 1 is a zero of the polynomial because a number a is called a zero when p(a)=0. Options B, C and D are incorrect: the degree is 2, the coefficients include 3 and −10, and the constant term is 7. Exam tip: to test a zero, substitute the given number and check whether the polynomial value is 0.

Substituting \(x=-3\), \(p(-3)=(-3)^2+6(-3)+9=9-18+9=0\). Therefore, 0 is the correct value, and \(-3\) is a zero of the polynomial. Choosing 9 usually results from omitting the term \(6(-3)\). In an exam, enclose a negative value in parentheses while substituting.

In \(x^2+2x+3\), the powers of \(x\) are 2, 1, and 0, all of which are non-negative integers; hence it is a polynomial. Its highest power is 2, so it is a quadratic polynomial, not a linear one. Option A has degree 1 and is linear, while option C contains \(x^{-1}\) and option D contains \(x^{1/2}\), so neither is a polynomial. Exam tip: in a polynomial, the variable can have only non-negative integer exponents such as 0, 1, 2, and so on.

The zero of a polynomial is the value of \(x\) that makes the polynomial equal to zero. Setting \(9x-27=0\) gives \(9x=27\), so \(x=3\). The value 9 is the coefficient of \(x\), not the zero. Exam tip: set a linear polynomial equal to zero and solve for \(x\).

The zero of a polynomial is the value of \(x\) for which \(p(x)=0\). Thus, \(4x+20=0\), so \(4x=-20\) and \(x=-5\). Therefore, option B is correct. \(-4\) is merely the negative of the coefficient and is not the zero. Exam tip: Set a linear polynomial equal to zero before solving for \(x\).

A zero of a polynomial is the value of \\(x\\) for which \\(p(x)=0\\). Thus, \\(cx+d=0\\) gives \\(cx=-d\\), and since \\(c\ne0\\), \\(x=-\frac{d}{c}\\). Option A misses the negative sign. For exams, remember that the zero of \\(ax+b\\) is \\(-\frac{b}{a}\\).

The (x^2)-term is (0x^2), so the coefficient is (0). A missing or zero term has coefficient (0).

The terms containing \(x^5\) and \(x^3\) are present. The constant term \(-2\) can be written as \(-2x^0\), so the power \(x^0\) is also present. There is no \(x^4\) term, which means its coefficient is 0. Exam tip: To identify a missing power, check whether its coefficient appears in the polynomial; a missing term has coefficient zero.

In a polynomial, powers of the variable are non-negative integers like (0,1,2,\ldots). This is the basic identification rule.

Substituting \(x=0\) gives \(p(0)=0^2-8(0)+13=13\), so the correct answer is 13. When \(x=0\), all terms containing \(x\) become zero, leaving only the constant term. Exam tip: \(p(0)\) is directly equal to the constant term of the polynomial.

\(p\left(\frac{2}{3}\right)=9\cdot\frac{4}{9}-8+4=0\). While substituting a fraction, square and multiply carefully.

The polynomial has no term containing \(x^3\), so the missing term is treated as \(0x^3\). Therefore, the coefficient of \(x^3\) is 0. The value -2 is the coefficient of \(x^2\), while 9 is the constant term. Exam tip: The coefficient of an absent power in a polynomial is taken as 0.

The coefficient of (x) is (-6) in the first polynomial and (9) in the second, so the sum is (3). Add coefficients of like powers only.

In (p(x)=x^3+1), the highest power is (3), so it is a cubic polynomial. Decide the type of a polynomial by its degree.

Substituting \(x=0\) gives \(p(0)=a(0)^3+b(0)^2+c(0)+d=d\). Since \(p(0)=-7\), the constant term is \(d=-7\). Option A has the incorrect sign. Exam tip: For any polynomial, \(p(0)\) equals its constant term.

For the degree to be (2), the coefficient of (x^4) must be (0), so (k-4=0) and (k=4). Check the coefficient of the highest power first.