Polynomials in one variable

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 \(-4\).

Since \(f(1)=0\) and \(f(2)=0\), the zeroes of the polynomial are 1 and 2. Therefore, \(f(x)=(x-1)(x-2)=x^2-3x+2\). Comparing this with \(x^2+px+q\), we get \(p=-3\) and \(q=2\). Hence, \(p+q=-3+2=-1\), so option A is correct. Exam tip: for a quadratic \(x^2+px+q\), the sum of zeroes is \(-p\) and their product is \(q\).

The sum of zeroes from the polynomial is (a) and the product is (b). Hence if the zeroes are (a,b), then (a+b=a) and (ab=b).

Using the difference-of-cubes identity, \(x^3-1^3=(x-1)(x^2+x+1)\). Hence, \(x-1\) is a linear factor of the polynomial. The other options are not factors because \(p(-1)=-2\), \(p(3)=26\), and \(p(-3)=-28\); thus, \(-1,3,-3\) are not zeroes. Exam tip: apply \(a^3-b^3=(a-b)(a^2+ab+b^2)\) when factoring a difference of cubes.

The polynomial can be written as \(p(x)=x^2-8x+16=(x-4)^2\). Therefore, \(p(4+t)=((4+t)-4)^2=t^2\). Option C gives only a linear term and misses the squaring step. In an exam, factoring into a perfect square is the quickest method here.

(p(x+1)=x^2-2x+3) and (p(x+1)-p(x)=2x-3). In such questions, substitute (x+1) carefully first.