Polynomials in one variable

Factoring the polynomial gives \(p(x)=(x-7)(x+4)\). Therefore, its zeroes are \(7\) and \(-4\). Their distance is \(|7-(-4)|=11\), so option A is correct. Exam tip: For two real zeroes, calculate the absolute value of their difference to obtain the distance.

For a quadratic polynomial \(ax^2+bx+c\), the product of its zeroes is \(\frac{c}{a}\). Here, the product is \(\frac{18}{2}=9\). If the other zero is \(\alpha\), then \(3\alpha=9\), giving \(\alpha=3\). Thus, the correct answer is 3. Exam tip: Find the product of the zeroes by dividing the constant term by the coefficient of \(x^2\).

(4x^2-4x+1=(2x-1)^2), so the zero (\frac{1}{2}) occurs twice. Perfect square form gives equal zeroes directly.

The polynomial factors as \(p(x)=(x^2-1)(x^2-4)\). Thus, \(x^2=1\) gives \(x=\pm1\), and \(x^2=4\) gives \(x=\pm2\). Therefore, the real zeroes are \((-2,-1,1,2)\). Option B is incorrect because \(x^2=4\) has roots \(\pm2\), not \(\pm4\). Exam tip: For a polynomial involving only even powers of \(x\), factor it by treating \(x^2\) as a single variable.

The direct answer is option A: \(a+b=-13\). For a monic cubic with zeroes \(\alpha,\beta,\gamma\), \(x^3+ax^2+bx+c=x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma\). Here the zeroes are 1, 3 and -4. Their sum is \(1+3-4=0\), so \(-a=0\), giving \(a=0\). Their pairwise-product sum is \(1\cdot3+3\cdot(-4)+(-4)\cdot1=3-12-4=-13\), so \(b=-13\). Therefore \(a+b=0+(-13)=-13\). Option A is correct because it gives -13. Option B gives 9, but neither the required coefficient calculation nor the result is 9. Option C gives -7, which comes from an incorrect combination of the roots. Option D gives 7 and is also based on an incorrect sign or product. The constant term checks the roots: \(-1\cdot3\cdot(-4)=-12\), matching the polynomial. Memory cue: for a monic cubic, the coefficient of \(x^2\) is the negative of the root sum, while the coefficient of \(x\) is the pairwise-product sum.

Factoring gives \(p(x)=x(x^2-9)=x(x-3)(x+3)\). Thus, the zeroes are \(x=0,3,-3\), which are three distinct real numbers. Therefore, the correct answer is 3. Exam tip: when counting distinct zeroes, count a repeated zero only once.

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.

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.

First, \(p(1)=2(1)^2+1-6=-3\) and \(p(-2)=2(-2)^2-2-6=0\). Therefore, \(2p(1)-p(-2)=2(-3)-0=-6\), so option A is correct. Exam tip: when substituting \(x=-2\), use parentheses because \((-2)^2=4\), not \(-4\).