Polynomials in one variable

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.

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).

Here (\alpha+\beta=\frac{10}{3}) and (\alpha\beta=\frac{7}{3}). Hence (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\frac{100}{9}-\frac{14}{3}=\frac{58}{9}).

The product of zeroes is (2\cdot(-3)=-6), and it equals (\frac{c}{a}). Pay special attention to the sign in products.

Let the two zeroes be \(t\) and \(t+2\). Their sum is \(6\), so \(t+(t+2)=6\), giving \(t=2\) and the other zero as \(4\). The product of the zeroes equals \(s\); hence, \(s=2\times4=8\). Exam tip: For \(x^2+bx+c\), the sum of the zeroes is \(-b\) and their product is \(c\).

(\alpha+\beta=\frac{5}{2}) and (\alpha\beta=-\frac{3}{2}). (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=-\frac{37}{6}).

For the polynomial to have degree 3, the coefficient of \(x^5\) must be zero so that the fifth-degree term disappears. Thus, \(k-3=0\), giving \(k=3\). The polynomial then becomes \(2x^3-x+9\), whose degree is indeed 3. Exam tip: set the coefficient of every term with a degree higher than the required degree to zero; for \(k=2\), the degree would remain 5.

(\alpha+\beta=9) and (\alpha\beta=20). ((\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)+4=20+18+4=42).

The sum of reciprocals is (\frac{\alpha+\beta}{\alpha\beta}). Here (\alpha+\beta=4) and (\alpha\beta=1), so the answer is (4).

Let the zeroes be \(\alpha=2+\sqrt{3}\) and \(\beta=2-\sqrt{3}\). Their sum is \(\alpha+\beta=4\), and their product is \(\alpha\beta=2^2-(\sqrt{3})^2=1\). A monic quadratic with zeroes \(\alpha\) and \(\beta\) is \(x^2-(\alpha+\beta)x+\alpha\beta\), so it is \(x^2-4x+1\). Option B has the wrong sign for the sum. Exam tip: for a monic quadratic, the coefficient of \(x\) is the negative of the sum of the zeroes, while the constant term is their product.

We can factor the polynomial as \(x^3-6x^2+12x-8=(x-2)^3\), using the identity \((a-b)^3=a^3-3a^2b+3ab^2-b^3\). Therefore, \(p(x)=0\) only when \(x=2\), and the factor \((x-2)\) occurs three times; hence the zero is 2 with multiplicity 3. Option B reverses the zero and its multiplicity. In an exam, first try to express a cubic as a perfect cube to identify repeated zeros quickly.

Here, \(a=3\), \(b=-12\), and \(c=15\). The discriminant is \(D=b^2-4ac=(-12)^2-4(3)(15)=-36<0\), so the quadratic polynomial has no real zeroes. Equivalently, \(p(x)=3[(x-2)^2+1]\), which is positive for every real \(x\). Thus option B is incorrect because equal real zeroes require \(D=0\). Exam tip: for a quadratic polynomial, \(D<0\) means that it has no real zeroes.

The direct answer is option A, 7 and 8. First find the zeroes of \\(p(x)=x^2-11x+30\\). It factors as \\(p(x)=(x-5)(x-6)\\), so its zeroes are 5 and 6. Now \\(q(x)=p(x-2)\\). To find a zero of q, set \\(p(x-2)=0\\). The input of p, namely x-2, must therefore equal one of p's zeroes: \\(x-2=5\\) or \\(x-2=6\\). Solving gives \\(x=7\\) or \\(x=8\\). Thus option A is correct. Option B, 3 and 4, shifts the zeroes in the wrong direction by subtracting 2. Option C, 5 and 6, gives the original p zeroes and ignores the replacement x-2. Option D, 2 and 11, uses unrelated numbers from the expression and is not obtained by solving. A useful cue: replacing x by x-a shifts every zero to the right by a.