Polynomials in one variable

For degree (2), the coefficient of (x^3) must be (0) and the coefficient of (x^2) must be non-zero. Both conditions hold when (m=0).

(p(-x)=3x^2+2x+1), so the sum is (6x^2+2). Odd-power terms change signs when (x) is replaced by (-x).

p(−2) = (−2)³ − 4(−2) = −8 + 8 = 0, and p(2) = 2³ − 4(2) = 8 − 8 = 0. Therefore, p(−2) + p(2) = 0 + 0 = 0. Also, p(x) is an odd polynomial, so p(−x) = −p(x), making the sum for opposite inputs zero. Exam tip: Check whether the polynomial is odd before calculating both values.

Completing the square gives \(p(x)=x^2+4x+5=(x+2)^2+1\). Since \((x+2)^2\geq 0\), the expression is smallest when \(x+2=0\), so \(x=-2\). Hence, option B is correct. Exam tip: write a quadratic in the form \(a(x-h)^2+k\); when \(a>0\), its minimum occurs at \(x=h\).

Completing the square gives \(p(x)=x^2-6x+10=(x-3)^2+1\). Since \((x-3)^2\geq 0\), we have \(p(x)\geq 1\), and the minimum value 1 occurs at \(x=3\). Option 3 is the \(x\)-coordinate of the vertex, not the minimum value. Exam tip: Rewrite a quadratic as \((x-a)^2+b\); when the square term has a positive coefficient, its minimum value is \(b\).

Since the constant term is \(9=3^2\), the perfect-square form must be \((x+3)^2=x^2+6x+9\). Therefore, \(k=6\). Although \((x-3)^2\) gives \(k=-6\), it is excluded because \(k>0\). In exams, identify the middle term of a square as \(2ab\).

In option A, \(x^2+10x+25=x^2+2(x)(5)+5^2=(x+5)^2\), so it is a perfect-square polynomial. In option B, the constant term is 20, whereas 25 is required to complete the square for \(x^2+10x\). In option C, the middle term is 5x instead of the required 10x, and option D has a negative constant term. In an exam, check the identities \(a^2+2ab+b^2=(a+b)^2\) and \(a^2-2ab+b^2=(a-b)^2\).

The polynomial can be written as \(p(x)=x^3-3x^2+3x-1=(x-1)^3\). Hence, \(p(2)=(2-1)^3=1\), so option B is correct. Option C incorrectly treats the input value 2 as the value of the polynomial. Exam tip: Look for the identity \(a^3-3a^2b+3ab^2-b^3=(a-b)^3\) before expanding and calculating term by term.

The governing concept is direct evaluation of a polynomial at specified values, followed by multiplication. For x = 1, p(1) = 2(1)³ + (1)² − 5(1) + 2 = 2 + 1 − 5 + 2 = 0. For x = −1, carefully preserve the odd-power sign: p(−1) = 2(−1)³ + (−1)² − 5(−1) + 2 = −2 + 1 + 5 + 2 = 6. Thus p(1)p(−1) = 0 × 6 = 0, making option C correct. Once one factor is zero, the product is zero regardless of the other factor. Options A, B, and D arise from substitution, sign, or multiplication errors. The corrected answer is C, not A.

To find the difference \\(p(2)-p(-1)\\), evaluate the polynomial at both inputs separately and then subtract the second result from the first. Careful handling of the negative value is important, especially for the cubic term. The two values are 12 and 6, so their difference is 6. Hence option D is correct.

For x=2, \\(p(2)=2(2)^3+(2)^2-5(2)+2=16+4-10+2=12\\). For x=-1, \\(p(-1)=2(-1)^3+(-1)^2-5(-1)+2=-2+1+5+2=6\\). Therefore \\(p(2)-p(-1)=12-6=6\\). The signs in \\((-1)^3\\) and \\(-5(-1)\\) must be handled carefully; confusing them can lead to another option.

Since \\(f(0)=q=6\\), the condition \\(f(2)=0\\) gives \\(2^2+2p+6=0\\). Thus, \\(4+2p+6=0\\), so \\(2p=-10\\) and \\(p=-5\\). Exam tip: use the value at \\(x=0\\) first to determine the constant term, then apply the second condition.

The equation \(p(0)=0\) directly means that \(x=0\) is a zero of the polynomial \(p(x)\). In fact, \(p(x)=x(x^2+ax+b)\), so \(x\) is a factor. This does not require \(a=0\) or \(b=0\); also, \(p(1)=1+a+b\), so 1 need not be a zero. Exam tip: If the constant term of a polynomial is zero, then zero is one of its zeros.

A number r is a zero of a polynomial p(x) when p(r)=0. To test x=0, substitute zero into each option. For x³−2x, the value is 0−0=0. For 5x²+x, it is 0+0=0. For x⁴, it is 0, so all three have zero as a zero. For x²+1, the value is 0²+1=1, not zero; therefore option D is the unique correct answer. The decisive feature is the constant term: at x=0, every positive-power term vanishes and the polynomial value equals its constant term. The first three expressions have constant term 0, whereas x²+1 has constant term 1.

To find \(p(x+1)\), replace every occurrence of \(x\) in the polynomial with the complete expression \((x+1)\): \(p(x+1)=(x+1)^2-2(x+1)+3=x^2+2x+1-2x-2+3=x^2+2\). Hence, option A is correct. Option D fails to combine the \(-2x\) term correctly. Exam tip: keep the substituted binomial in parentheses, especially when it is squared.

To find \(p(2x)\), replace every \(x\) in the polynomial with \(2x\): \(p(2x)=2(2x)^2+3(2x)-4=8x^2+6x-4\). Hence, option A is correct. In option B, the term \(2(2x)^2\) has been incorrectly simplified as \(4x^2\). Exam tip: when substituting an expression for a variable, apply the power to the entire substituted expression.

First calculate the fixed value p(1): p(1)=1²−4(1)+1=1−4+1=−2. Now subtract this value from the general expression: p(x)−p(1)=(x²−4x+1)−(−2)=x²−4x+1+2=x²−4x+3. Therefore option A is correct. The negative sign before p(1) is important: subtracting −2 is the same as adding 2. Option C is merely the original polynomial and ignores the subtraction. Option B would result from subtracting 2 rather than subtracting −2, and option D changes the sign of the linear term without any algebraic basis. The original option list and key were inconsistent; option A has been corrected to the actual result.

The highest power is (3), so the polynomial is cubic. In exams, always check the highest non-zero exponent.

For a zero \(\alpha\) of a polynomial, \(p(\alpha)=0\). Thus, \(p(2)=4k-10+6=0\), which gives \(4k-4=0\) and hence \(k=1\). Therefore, option A is correct. In such questions, substitute the given zero into the polynomial and set the result equal to zero.

For a quadratic polynomial \(ax^2+bx+c\), the sum of its zeroes is \(-\frac{b}{a}\). Here, the sum of the zeroes is \(3+4=7\), while the given polynomial shows that this sum is \(m+3\). Thus, \(m+3=7\), giving \(m=4\). As a check, the product of the zeroes is \(3\times4=12\), which agrees with the constant term. Exam tip: In the form \(x^2-Sx+P\), the sum of the zeroes can be read directly as \(S\).

For a quadratic polynomial \(ax^2+bx+c\), the product of its zeroes is \(\frac{c}{a}\). Here, \(a=2\) and \(c=3\), so the product is \(\frac{3}{2}\). Option C, \(-\frac{7}{2}\), is related to the sum of the zeroes, \(-\frac{b}{a}\), not their product. Exam tip: remember that the sum of zeroes is \(-\frac{b}{a}\), while their product is \(\frac{c}{a}\).