Polynomials in one variable

By Vieta’s formulas, \(\alpha+\beta=8\), since the coefficient of \(x\) is \(-8\). We are also given that \(\alpha-\beta=2\). Adding the two equations gives \(2\alpha=10\), so \(\alpha=5\) and \(\beta=3\). For \(x^2-8x+k\), the product of the zeroes is \(\alpha\beta=k\). Hence, \(k=5\times3=15\). Exam tip: In a monic quadratic \(x^2+bx+c\), the product of the zeroes is \(c\), so 16 is not correct.

For a quadratic polynomial \(ax^2+bx+c\), the sum of its zeroes is \(-\frac{b}{a}\). Here, \(a=k\) and \(b=6\), so \(-\frac{6}{k}=-3\). Thus, \(6=3k\), giving \(k=2\). Option B results from a sign error; if \(k=-2\), the sum would be \(3\). Exam tip: use the coefficient formula \(-\frac{b}{a}\) directly for the sum of zeroes.

By Vieta’s formula, the sum of the zeroes is \(2r+3r=5r\). Thus, \(5r=10\), giving \(r=2\). Their product is \((2r)(3r)=6r^2=6\times 2^2=24\). For the monic quadratic \(x^2-10x+q\), the product of the zeroes equals the constant term \(q\), so \(q=24\). Exam tip: for \(x^2+bx+c\), the product of the zeroes is always \(c\).

Since the zeroes are −1, −2, and −4, the polynomial can be written as (x + 1)(x + 2)(x + 4). Expanding gives x³ + 7x² + 14x + 8. Thus, a = 7 and b = 14, so a + b = 21. Options 7 and 14 represent a and b separately, not their sum. Exam tip: For a cubic polynomial, compare the coefficients of x² and x after forming the product of the corresponding linear factors.

(\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)). Here (\alpha\beta=-10) and (\alpha+\beta=3), so the value is (-30).

If the zeroes are 0, 2, and −3, the polynomial is formed as \(x(x-2)(x+3)\). Expanding it gives \(x(x^2+x-6)=x^3+x^2-6x\), so option A is correct. Option B factors as \(x(x-3)(x+2)\), giving different zeroes. Exam tip: form one factor \((x-\text{zero})\) for each zero and then expand.

For the quadratic polynomial \(x^2+mx+n\), the sum of its zeros is \(-m\) and their product is \(n\). Here, the sum is \(0+5=5\), so \(-m=5\), giving \(m=-5\). Their product is \(0\times5=0\), so \(n=0\). Therefore, \(m+n=-5+0=-5\). Exam tip: In the standard form \(x^2+mx+n\), remember that the sum of zeros is \(-m\) and the product is \(n\).

For the quadratic polynomial, \(\alpha+\beta=11\) and \(\alpha\beta=30\). Therefore, \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=11^2-4(30)=121-120=1\). Hence, option A is correct. Exam tip: To find the square of the difference of the zeroes, use their sum and product; taking only \((\alpha+\beta)^2\) gives 121, not the required value.

Factoring gives \(4x^2-12x+9=(2x-3)^2\). Hence \(2x-3=0\), so \(x=\frac{3}{2}\), and this zero is repeated; therefore, both zeroes are \(\frac{3}{2}\). Option C is incorrect because it treats the zeroes as two distinct values. Exam tip: A quadratic polynomial that is a perfect square, or has discriminant zero, has equal zeroes.

By the Factor Theorem, if \(p(a)=0\), then \((x-a)\) is a factor of the polynomial. Here, \(p(1)=1-7+6=0\) and \(p(2)=8-14+6=0\). Therefore, both \((x-1)\) and \((x-2)\) are factors. Exam tip: To test whether \((x-a)\) is a factor, evaluate \(p(a)\) directly.

By the factor theorem, if \(x+2\) is a factor, then \(p(-2)=0\). Therefore, \((-2)^3+m(-2)^2-4(-2)-4=0\), which gives \(-8+4m+8-4=0\). Hence \(4m-4=0\), so \(m=1\). Exam tip: for a factor \(x-a\), substitute \(x=a\); thus, for \(x+2=x-(-2)\), substitute \(x=-2\).

The conditions give (2a+b=-1) and (-a+b=5), so (a=-2), (b=3), and (a+b=1). When two values are given, form two equations.

Here (\alpha+\beta=6) and (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta). So (20=36-2k), giving (k=8).

For degree (4), the coefficient of (x^5) must be (0) and the coefficient of (x^4) must be non-zero. Both conditions hold for (m=-2).

Substituting (x=1) gives (p(1)=k(1)^3-2(1)^2+5(1)-4=k-1). Hence, (k-1=7), so (k=8). The distractor 7 results from overlooking the net contribution of the constant terms, (-2+5-4=-1). Exam tip: whenever (p(a)) is given, substitute x=a first and then solve the resulting equation.

The coefficients of \(x\) in the two polynomials are \(2\) and \(-6\), respectively. Therefore, the coefficient of \(x\) in their sum is \(2+(-6)=-4\). The complete sum is \(3x^3-4x^2-4x+4\). Exam tip: when adding polynomials, combine coefficients only for terms with the same power.

The direct answer is option C: the degree is \(3\). Add like powers carefully: \(p(x)+q(x)=(5x^4-2x^2+1)+(-5x^4+3x^3+x-6)\). The fourth-degree terms cancel: \(5x^4-5x^4=0\). The remaining polynomial is \(3x^3-2x^2+x-5\). Its highest power with a non-zero coefficient is 3, so its degree is 3. Option A, 1, is wrong because the \(x^3\) and \(x^2\) terms remain. Option B, 2, ignores the non-zero \(3x^3\) term. Option C is correct. Option D, 4, would be correct only if the fourth-degree terms did not cancel. Never decide the degree from the original expressions before adding them. Memory cue: combine first, then read the highest surviving power.

The direct answer is option C, 7. For non-zero polynomials, the degree of a product equals the sum of the degrees. Let the leading terms be \\(ax^5\\) and \\(bx^2\\), with a and b non-zero. Their product begins with \\(abx^7\\), and ab is non-zero, so the degree is 7. Thus \\(\\deg(pq)=5+2=7\\), making option C correct. Option A, 3, could result from subtracting the degrees, but subtraction is not the rule for multiplication. Option B, 5, is only the degree of p and leaves q out. Option D, 10, comes from multiplying 5 and 2; degree multiplication is not used here. The wording says the product is non-zero, which removes any concern about a zero polynomial. Exam cue: for a product of non-zero polynomials, add the highest powers.

If a number is a zero of a polynomial, substituting that number makes the polynomial equal to zero. Since -2 is a zero of p(x)=x^2+kx-8, use p(-2)=0. Substitution gives (-2)^2+k(-2)-8=0, so 4-2k-8=0. Simplifying, -2k-4=0, which means -2k=4 and k=-2. Therefore choice B is correct.

The sign of the middle term must be handled carefully: kx becomes k(-2)=-2k. One may also check the result directly. With k=-2, p(x)=x^2-2x-8, and p(-2)=4+4-8=0. This confirms that -2 is indeed a zero and that the selected value is consistent.

Substituting x = 2 gives p(2) = 3(2³) − 8(2²) + 7(2) − 2 = 24 − 32 + 14 − 2 = 4. Therefore, option B is correct. Exam tip: evaluate the powers first, then multiply by their coefficients and perform the operations in order; adding the coefficients alone would be incorrect.

Substituting \(x=1\) gives \(p(1)=1^3-5(1)^2+m(1)+8=1-5+m+8=m+4\). Since \(p(1)=0\), we require \(m+4=0\), so \(m=-4\). For example, using \(m=4\) gives \(p(1)=8\), not zero. Exam tip: to determine a parameter from a given zero of a polynomial, substitute the zero directly and solve the resulting equation.

Substituting \(x=-2\), \(p(-2)=(-2)^4-5(-2)^2+4=16-5(4)+4=16-20+4=0\). Since the powers are even, \((-2)^4=16\) and \((-2)^2=4\). Option B results from an incomplete calculation, so it is incorrect. In exams, substitute the value first, evaluate powers, and then perform multiplication and subtraction.

Since \(p(2)=0\) and \(p(4)=0\), 2 and 4 are the zeros of the polynomial. Therefore, \(p(x)=(x-2)(x-4)=x^2-6x+8\). Comparing coefficients gives \(a=-6\) and \(b=8\), so \(a+b=-6+8=2\). Option 3 is not correct; the coefficients must be compared after factoring. Exam tip: if the zeros of a monic quadratic are \(r\) and \(s\), then \(p(x)=x^2-(r+s)x+rs\).

Since \(-1\) and \(5\) are the zeroes of \(p(x)\), we can write \(p(x)=(x+1)(x-5)=x^2-4x-5\). Comparing this with \(x^2+ax+b\), the constant term is \(b=-5\). Therefore, option A is correct. Exam tip: For a monic quadratic polynomial, the product of its zeroes equals its constant term.

Since \(a=0\) and \(b=0\), the terms \(ax^4\) and \(bx^3\) disappear. Because \(c\ne 0\), the term \(cx^2\) definitely remains, giving \(p(x)=cx^2+d\). Therefore, the degree is \(2\). The constant term \(d\) does not change the degree. Exam tip: the degree of a polynomial is the exponent of the highest-power term with a non-zero coefficient.