Polynomials in one variable

For (p(a)=0ig), we require (a^2-9=0ig), so (a^2=9ig) and (a=\pm3ig). Since the question asks for the positive value, the correct answer is 3. Note that (a=-3ig) is also a zero of the polynomial, but it is not positive. Exam tip: Factor the expression as (x^2-9=(x-3)(x+3)ig) to find its zeros quickly.

The polynomial can be factorised as \(p(x)=(2x-3)^2\). Thus, \(p(x)=0\) when \(2x-3=0\), giving \(x=\frac{3}{2}\). Indeed, \(p\left(\frac{3}{2}\right)=9-18+9=0\), whereas the other options do not make the polynomial zero. Exam tip: For a quadratic polynomial, try factorisation before substituting every option.

The conditions give (a+b=-7) and (4a-2b=2), so (a=-2), (b=-5), and (a-b=3). The correct value is (3).

(p(1)=1+a+b+4=0) and (p(-1)=-1+a-b+4=0); adding gives (2a+8=0), so (a=-4). Add signs carefully.

By the Factor Theorem, if x − 2 is a factor of p(x), then p(2) = 0. Therefore, 2³ + k(2²) − 4(2) − 12 = 0, giving 8 + 4k − 8 − 12 = 0. Thus, 4k − 12 = 0 and k = 3. Option B results from a sign error. Exam tip: whenever x − a is a factor, substitute x = a and set the polynomial equal to zero.

The polynomial has no visible x²-term, which means its x²-term is 0x². Therefore, the coefficient of x² is 0. Here, −3 is the coefficient of x³ and 5 is the coefficient of x. Exam tip: The coefficient of a missing term in a polynomial is taken as 0.

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.

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.

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.

The direct answer is C: the polynomial is cubic, not linear. The degree of a polynomial is the greatest exponent of the variable whose coefficient is not zero. In \(p(x)=ax^3+bx^2+cx+d\), the condition \(a\ne0\) guarantees that the term \(ax^3\) is present, so the highest power is exactly 3. Therefore the polynomial has degree 3 and is called a cubic polynomial. Option A is wrong because degree 1 describes a linear polynomial, such as \(mx+n\). Option B is wrong because degree 2 describes a quadratic polynomial; the \(x^2\) term is not the highest term here. Option C is correct because the non-zero leading term has power 3. Option D is irrelevant: \(d\) is only the constant term and does not decide the degree. Always inspect the largest 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 \(\frac{c}{a}\).

p(2)=2^3-4(2^2)+2+6=8-16+2+6=0. Therefore, 2 is a zero of the polynomial. Option B is incorrect because a number is a zero only when the polynomial evaluates to zero at that number. Exam tip: if p(a)=0, then a is a zero and (x-a) is a factor of the polynomial.

Using the given zeroes, the polynomial can be written as \((x-1)(x-2)(x-3)\). On expansion, this becomes \(x^3-6x^2+11x-6\). Therefore, \(p=-6\) and \(q=11\), so \(p+q=-6+11=5\). The value \(-5\) results from subtracting instead of adding the coefficients. In an exam, form the factors from the zeroes, expand them, and compare coefficients of like powers.