Polynomials in one variable

Factor the polynomial as \(p(x)=(x-4)(x+2)\), so its zeroes are \(4\) and \(-2\). For \(p(x+3)=0\), we must have \(x+3=4\) or \(x+3=-2\), giving \(x=1\) or \(x=-5\). Therefore, the zeroes are \((-5,1)\). Exam tip: the zeroes of \(p(x+a)\) are obtained by subtracting \(a\) from each zero of \(p(x)\).

A non-zero polynomial of degree 4 can have at most 4 zeroes. It may have fewer zeroes, but never more than its degree. Exam tip: maximum zeroes equal the degree.

A polynomial of degree at most (2) can have three distinct zeroes only if it is the zero polynomial. A non-zero polynomial cannot have more distinct zeroes than its degree.

Let the zeroes be t and 4t. Their product equals the constant term, so t(4t)=36, giving 4t^2=36. Since the zeroes are positive, t=3. Their sum is therefore 3+12=15. For x^2-kx+36, the sum of the zeroes is k, so k=15. Exam tip: For x^2+bx+c, the sum of the zeroes is -b and their product is c.

For ax² + bx + c, if the zeroes are α and β, then α + β = −b/a and αβ = c/a. In this polynomial, a = 2, b = −7, and c = 5. Hence the sum is −(−7)/2 = 7/2, while the product is 5/2. The required difference is sum minus product: 7/2 − 5/2 = 2/2 = 1. Therefore option A is correct. Option B would result from an arithmetic error, option C confuses the reciprocal coefficient relationship, and option D does not follow from either standard relation. The calculation uses the coefficient formulas directly, so solving for the individual zeroes is unnecessary.

Since
(p(2)=0)
, substitute
(x=2)
into the polynomial:
2^3+2m-10=0
. Thus,
8+2m-10=0
, so
2m=2
and
(m=1)
. Therefore, option A is correct. Exam tip: To evaluate a polynomial at a given value, substitute that value directly for
(x)
.

The polynomial can be written as \(p(x)=x^2+4x+4=(x+2)^2\). Substituting \(x=-2+h\) gives \(p(-2+h)=((-2+h)+2)^2=h^2\). Option D results from confusing a squared expression with a linear term. Exam tip: Factor or rewrite a polynomial as a perfect square before substituting whenever possible.

The sum of zeroes is (-3), so (p=3). The pairwise product sum is (-3+5-15=-13), so (q=-13) and (p+q=-10).

(\alpha+\beta=12) and (\alpha\beta=35). (\frac{1}{\alpha-1}+\frac{1}{\beta-1}=\frac{\alpha+\beta-2}{\alpha\beta-\alpha-\beta+1}=\frac{10}{24}=\frac{5}{12}).

In a polynomial, every exponent of \(x\) must be a non-negative integer. In option B, the greatest exponent is 4, so its degree is 4. A, C and D contain negative or fractional exponents. Exam tip: check exponents first.

The sum is (-\frac{b}{a}=-\frac{-20}{5}=4). Even if a common factor is removed first, the sum remains the same.

A factor contributes real zeroes only when it becomes zero for real values of \(x\). Here, \(x^2+4=0\) gives \(x^2=-4\), which has no real solution; its complex roots are \(x=\pm 2i\). Therefore, this factor contributes no real zero. The values \(2\) and \(-2\) come from the factor \(x^2-4\), not from \(x^2+4\). Exam tip: set each factor equal to zero and check whether its roots are real.

Substituting \(x=1\) in the polynomial gives \(p(1)=1+a+a=1+2a\). Since \(p(1)=0\), we get \(1+2a=0\), so \(2a=-1\) and \(a=-\frac{1}{2}\). Therefore, option A is correct. Option C has the wrong sign; if \(a=\frac{1}{2}\), then \(p(1)=2\), not zero. Exam tip: When a polynomial value is specified, substitute the given value of \(x\) directly and simplify the resulting equation.

Completing the square gives \(p(x)=x^2-2x+k=(x-1)^2+k-1\). Since \((x-1)^2\geq 0\), the minimum value occurs at \(x=1\) and is \(k-1\). Therefore, \(k-1=3\), so \(k=4\). Exam tip: rewrite a quadratic as a square plus a constant to identify its minimum or maximum value quickly.

Factoring gives \(p(x)=(x-1)(x-3)\). The product is negative between its two roots, so \(p(x)<0\) when \(1<x<3\). Among the given choices, only \(x=2\) lies in this interval. Hence, option A is correct. Exam tip: For a quadratic with a positive leading coefficient, the expression is negative between two distinct real roots and positive outside them.

Grouping the terms gives \(p(x)=x^2(x-3)-4(x-3)\). Therefore, \(p(x)=(x-3)(x^2-4)=(x-3)(x-2)(x+2)\), so option A is correct. In option B, the first factor is incorrectly written as \(x+3\), whereas grouping clearly gives \(x-3\). In an exam, first extract the common binomial and then apply \(a^2-b^2=(a-b)(a+b)\).

First, \(p(x)=x^2+5x+6=(x+2)(x+3)\), so its zeroes are \(-2\) and \(-3\). Therefore, \(r(x)=p(-x)=(-x)^2+5(-x)+6=x^2-5x+6=(x-2)(x-3)\), giving the zeroes \(2\) and \(3\). Option B lists the zeroes of the original polynomial, but replacing \(x\) by \(-x\) changes the signs of the zeroes. Exam tip: In \(p(-x)\), substitute \(-x\) for every occurrence of \(x\), then factor the resulting polynomial.

(x^2-7x+12=(x-3)(x-4)), so the zeroes are (3) and (4). The (x)-axis points are ((3,0)) and ((4,0)).

A quadratic polynomial ax² + bx + c has no real zero when its discriminant b² − 4ac is less than 0. Here, the discriminant is (2)² − 4(1)(c) = 4 − 4c. Thus, 4 − 4c < 0 gives c > 1. For c = 1, the discriminant is zero and the polynomial has one real zero, so option B is incorrect. Exam tip: Always check the discriminant to determine the number of real zeroes of a quadratic polynomial.

Let α and β be the zeroes of p(x). Since p(x)=x^2-6x+8 is monic, Vieta’s relations give α+β=6 and αβ=8. The required zeroes are α^2 and β^2. Their sum is α^2+β^2=(α+β)^2-2αβ=6^2-2(8)=36-16=20. Their product is α^2β^2=(αβ)^2=8^2=64. A monic quadratic with zeroes having sum S and product P is x^2-Sx+P. Substituting S=20 and P=64 gives x^2-20x+64, so option A is correct. Option B is merely the original polynomial. Option C uses an incorrect transformed sum and product, while D has product 64 but the wrong sum. The calculation can also be checked from the actual zeroes 2 and 4, whose squares are 4 and 16; their polynomial is (x-4)(x-16)=x^2-20x+64.