Polynomials in one variable

The governing concept is substitution of -x and cancellation of odd-power terms. Starting with p(x)=5x^2-4x+3, we get p(-x)=5(-x)^2-4(-x)+3=5x^2+4x+3. The square term is unchanged because (-x)^2=x^2, while the linear term changes sign. Adding gives p(x)+p(-x)=(5x^2-4x+3)+(5x^2+4x+3)=10x^2+6. Thus option A is correct. The terms -4x and +4x cancel, while the two quadratic terms and the two constants add. Option B incorrectly keeps the linear term, C omits one quadratic contribution, and D incorrectly removes the quadratic terms. This also illustrates that the sum of a polynomial and its reflection in x is even.

\(p(2)=3(2)^3-12(2)=24-24=0\), while \(p(-2)=3(-2)^3-12(-2)=-24+24=0\). Hence \(p(-2)+p(2)=0+0=0\), so option B is correct. Remember that the cube of a negative number is negative, whereas \(-12(-2)\) is positive. In an exam, evaluate both values separately and check the signs carefully.

Completing the square gives \(p(x)=x^2-12x+40=(x-6)^2+4\). Since \((x-6)^2\geq 0\), its least value is 0 when \(x=6\). Therefore, the polynomial has minimum value 4 at \(x=6\). As an exam tip, writing a quadratic in the form \(a(x-h)^2+k\) immediately gives the vertex's x-coordinate as \(h\) when \(a>0\).

Completing the square gives \(p(x)=x^2+14x+53=(x+7)^2+4\). Since \((x+7)^2\geq 0\), we have \(p(x)\geq 4\), and equality occurs at \(x=-7\). Therefore, the minimum value is 4, so option B is correct. The nearby distractor 2 does not follow from the correct completed-square form. Exam tip: when a quadratic is written as \((x-h)^2+k\) with a positive leading coefficient, its minimum value is \(k\).

A perfect-square form of \(x^2+kx+25\) must be \((x\pm5)^2\). Since \(k<0\), use \((x-5)^2=x^2-10x+25\), giving \(k=-10\). The form \((x+5)^2\) gives \(k=10\), but it does not satisfy the stated condition. Exam tip: in \((x+a)^2=x^2+2ax+a^2\), the coefficient of the middle term is \(2a\).

Recognising the identity gives \(p(x)=(x-2)^3\), since \((x-2)^3=x^3-6x^2+12x-8\). Therefore, \(p(5)=(5-2)^3=3^3=27\). The value 32 can result from an error while evaluating the cube. In an exam, first check whether a cubic polynomial matches a standard identity such as \((x-a)^3\).

\(p(2)=2^3-3(2)^2-4(2)+12=8-12-8+12=0\), and \(p(-2)=(-2)^3-3(-2)^2-4(-2)+12=-8-12+8+12=0\). Therefore, \(p(2)\times p(-2)=0\times0=0\), so option B is correct. In an exam, remember that finding even one zero factor is sufficient to conclude that the entire product is zero.

The governing concept is direct evaluation of a polynomial at specified values, with careful attention to signs. At x=2, p(2)=2^3-2^2-4(2)+4=8-4-8+4=0. At x=-2, p(-2)=(-2)^3-(-2)^2-4(-2)+4=-8-4+8+4=0. Therefore p(2)-p(-2)=0-0=0, so option A is correct. The negative input must be enclosed in parentheses: (-2)^2 is 4, and -4(-2) is +8. A mistaken sign in either term can lead to options B, C, or D. In fact, the polynomial factors as (x-1)(x-2)(x+2), which independently confirms that both 2 and -2 are zeroes and that their difference is zero.

Since \(P(0)=q=12\), substituting \(x=3\) in \(P(x)=x^2+px+q\) gives \(9+3p+12=0\). Thus, \(3p=-21\) and \(p=-7\), so option A is correct. Exam tip: use the value at \(x=0\) first to determine the constant term, then substitute the second given value to find the unknown coefficient.

Substituting \(x=0\), we get \(p(0)=0^5+a\cdot0^3+b=b\). Hence the condition \(p(0)=0\) necessarily implies \(b=0\). The value of \(a\) is unrestricted; for example, \(a=2\) also satisfies the condition. Also, \(p(1)=1+a\), which is not zero for every possible value of \(a\). Since the coefficient of \(x^5\) is 1, the polynomial has degree 5, not 3. Exam tip: To find \(p(0)\), substitute zero; the result is the constant term.

Substituting (x=0) in (4x^3-7x) gives (0), and it is not the zero polynomial. For (x=0), the constant term must be (0).

To find \(p(x+3)\), replace every occurrence of \(x\) in the polynomial by the complete expression \((x+3)\): \(p(x+3)=(x+3)^2-2(x+3)+4=x^2+6x+9-2x-6+4=x^2+4x+7\). Hence, option A is correct. Option B results from simplifying the constant terms incorrectly. Exam tip: when substituting an expression for \(x\), place the entire expression in parentheses.

To find \(p(3x)\), replace every occurrence of \(x\) in the polynomial with the complete expression \(3x\): \(p(3x)=4(3x)^2-3(3x)+2=36x^2-9x+2\). In option B, \(4(3x)^2\) is incorrectly treated as \(12x^2\), whereas \((3x)^2=9x^2\). Exam tip: when substituting an expression, apply the exponent to the entire parenthesized expression.

First evaluate \(p(3)\): \(p(3)=3^2-8(3)+15=9-24+15=0\). Therefore, \(p(x)-p(3)=p(x)-0=x^2-8x+15\), so option A is correct. Option B results from the common error of taking \(p(3)=3\) instead of evaluating the polynomial. In an exam, substitute the given value into every occurrence of the variable before simplifying.

The conditions give (a+b=-21) and (4a-b=11), so (a=-2), (b=-19), and (a-b=17). When two zeros are given, form two equations.

(p(x+1)=2x^2-3x-2) and (p(x-1)=2x^2-11x+12), so the difference is (8x-14). Substitute the whole expression in place of (x).

By the factor theorem (p(3)=0), so (27-45+3a+12=0) and (a=2). In exams, substitute the zero from the given factor directly.

By the remainder theorem (p(-1)=7), so (-2+k+11+5=7). This gives (k=-7), so none of the listed options is correct.

The sum (4+m) and product (4m) match the given polynomial. Therefore the statement is true for every real (m).

For equal zeroes, (r^2-36=0), so (r=\pm6). For negative zeroes the sum must be positive (6), hence (r=6).