Polynomials in one variable

For a quadratic ax²+bx+c, equal zeroes occur precisely when its discriminant is zero. The discriminant is Δ=b²−4ac. In x²−9x+k, a=1, b=−9, and c=k, so Δ=(−9)²−4(1)(k)=81−4k. Setting this equal to zero gives 81−4k=0, hence 4k=81 and k=81/4. Therefore option C is correct. Option B is the square of 9 but omits the factor 4 from the discriminant formula. Option A and option D arise from incorrect rearrangement or division. The equal-root condition is the governing idea; no numerical solving of the roots is needed, although the resulting repeated root would be 9/2.

p(2)=2^3-4(2^2)+2+6=8-16+2+6=0. Therefore, 2 is a zero of the polynomial. Option B is incorrect because a number is a zero only when the polynomial evaluates to zero at that number. Exam tip: if p(a)=0, then a is a zero and (x-a) is a factor of the polynomial.

Using the given zeroes, the polynomial can be written as \((x-1)(x-2)(x-3)\). On expansion, this becomes \(x^3-6x^2+11x-6\). Therefore, \(p=-6\) and \(q=11\), so \(p+q=-6+11=5\). The value \(-5\) results from subtracting instead of adding the coefficients. In an exam, form the factors from the zeroes, expand them, and compare coefficients of like powers.

Using the given zeroes, the polynomial can be written as \((x-1)(x-2)(x-3)\). On expansion, this becomes \(x^3-6x^2+11x-6\). Comparing coefficients gives \(p=-6\) and \(q=11\), so \(p+q=-6+11=5\). Exam tip: For a monic cubic polynomial, compare the coefficients of \(x^2\) and \(x\) directly after forming the product of the corresponding linear factors.

For zeroes \(\alpha=\frac{2}{3}\) and \(\beta=-\frac{5}{4}\), their sum is \(-\frac{7}{12}\) and their product is \(-\frac{5}{6}\). For a quadratic polynomial \(ax^2+bx+c\), the sum of zeroes is \(-\frac{b}{a}\) and their product is \(\frac{c}{a}\). Option A gives exactly these two values, so it is correct. In option B, the sign of the middle term is wrong, which changes the sum of the zeroes. Exam tip: calculate the sum and product of the given zeroes first, then compare them with the options.

For a quadratic polynomial \(x^2+ax+b\), the sum of the zeroes is \(-a\) and their product is \(b\). Here, \(4+(-7)=-3\), so \(-a=-3\) and hence \(a=3\). Also, \(b=4\times(-7)=-28\). Therefore, \(a-b=3-(-28)=31\), so option B is correct. Exam tip: In \(x^2+px+q\), the sum of the zeroes is \(-p\), while their product is \(q\).

The degree of a polynomial is determined by the highest power whose coefficient is nonzero. A term with coefficient zero is actually zero and does not contribute to the polynomial or its degree. In option A, the apparent fifth-degree term is \\(0x^5\\), so it disappears. The remaining highest nonzero power is \\(x^4\\), giving degree 4.

Option A can be viewed as \\(7x^4-3x+1\\), because \\(0x^5=0\\). Its highest nonzero exponent is 4. Option B has degree 3 after removing \\(0x^6\\); option C has degree 2; and option D, the nonzero constant 9, has degree 0. Therefore only option A has degree 4.

\(p(3)=3^2-5(3)+6=9-15+6=0\) and \(p(2)=2^2-5(2)+6=4-10+6=0\). Therefore, \(p(3)-p(2)=0-0=0\), so option A is correct. Option C, \(-1\), can result from a common subtraction or substitution error. Exam tip: when evaluating a polynomial, substitute the value for the variable carefully in every term before simplifying.

For a quadratic polynomial \(ax^2+bx+c\), the sum of its zeroes is \(-\frac{b}{a}\). Here, \(a=2\) and \(b=k\), so \(-\frac{k}{2}=3\), which gives \(k=-6\). Exam tip: use \(-b/a\) for the sum of zeroes; \(c/a\) gives their product, not their sum.

For a quadratic polynomial \(ax^2+bx+c\), the product of its zeroes is \(\frac{c}{a}\). Here, \(a=3\) and \(c=m\), so the product is \(\frac{m}{3}\). Therefore, \(\frac{m}{3}=\frac{4}{3}\), which gives \(m=4\). Exam tip: remember that the sum of zeroes is \(-\frac{b}{a}\), while their product is \(\frac{c}{a}\).

The governing concept is factorisation of a quadratic polynomial and the relation between factors and zeroes. We factor the expression as x^2 - 6x + 9 = (x - 3)^2. Therefore, the equation (x - 3)^2 = 0 gives x = 3, counted twice. Hence the two zeroes are equal, so option B is correct. Option A is false because there are not two different roots. Option C is false because 3 is a real zero. Option D is false because the product of the zeroes is 3 × 3 = 9, not 6. The repeated factor also confirms that the quadratic has a repeated zero.

The governing concept is the factor theorem together with expansion of three linear factors. A cubic having factors (x-a), (x-b), and (x-c) can be written as (x-a)(x-b)(x-c), apart from a non-zero constant multiplier. First, (x-a)(x-b) = x^2-(a+b)x+ab. Multiplying by (x-c) gives x^3-(a+b+c)x^2+(ab+bc+ca)x-abc. Thus option A has exactly the required factors and is a possible monic cubic. Option B has incorrect signs, while options C and D place the symmetric sums with incorrect powers or signs. Substituting x=a, b, or c into option A gives zero, confirming each factor.

Factoring the polynomial gives \(x^3-6x^2+11x-6=(x-1)(x-2)(x-3)\). Setting each factor equal to zero gives \(x=1,2,3\), so option A is correct. The values in option C do not satisfy the required sum and product relationships for this cubic polynomial. Exam tip: For likely integer zeroes, substitute each candidate directly into the polynomial to verify it.

By the factor theorem, if \(p(a)=0\), then \(x-a\) is a factor of \(p(x)\). Here, \(p(2)=2^3-3(2)^2-4(2)+12=8-12-8+12=0\), so \(x-2\) is a factor. In fact, \(p(x)=(x-2)(x-3)(x+2)\); the other given options are not factors. Exam tip: to test a possible linear factor \(x-a\), calculate \(p(a)\) directly.

By the factor theorem, if \((x+1)\) is a factor, the polynomial must be zero at \(x=-1\). Thus, \(2(-1)^3+k(-1)^2-5(-1)+2=0\), giving \(-2+k+5+2=0\), so \(k+5=0\) and \(k=-5\). The value 5 is a close distractor caused by a sign error; remember that \(x+1=x-(-1)\), so substitute \(-1\), not 1.

By the Factor Theorem, if (x+1) is a factor of P(x), then P(-1)=0. Thus, P(-1)=2(-1)^3+k(-1)^2-5(-1)+2=-2+k+5+2=k+5. Hence, k+5=0, giving k=-5. Exam tip: for a factor of the form (x-a), substitute x=a; therefore, (x+1)=(x-(-1)) requires substituting x=-1.

By the remainder theorem, the remainder when p(x) is divided by x - a is p(a). Here, a = 1, so p(1) = 1^3 - 5(1)^2 + 8(1) - 4 = 1 - 5 + 8 - 4 = 0. Therefore, the remainder is 0. Exam tip: For a divisor of the form x - a, substitute a directly into the polynomial instead of performing long division.

By the Remainder Theorem, the remainder when \(p(x)\) is divided by \(x-a\) is \(p(a)\). Since \(x+2=x-(-2)\), substitute \(x=-2\): \(p(-2)=3(-2)^3-2(-2)^2+(-2)+7=-24-8-2+7=-27\). Therefore, option A is correct. Exam tip: Rewrite the divisor as \(x-a\) before substituting, so the sign of \(a\) is not missed.

By the Remainder Theorem, when a polynomial \(p(x)\) is divided by \(x-a\), the remainder is \(p(a)\). Here, \(a=4\) and the given remainder is \(7\), so \(p(4)=7\). The value \(0\) would apply if \(x-4\) were a factor of \(p(x)\). Exam tip: For a divisor of the form \(x-a\), substitute \(x=a\) in the polynomial.

(x^2+1) is not linear because its degree is (2). Over real numbers, it is not a linear factor.