Linear polynomials

Given p(x)=7x+c, substituting x=1 gives p(1)=7(1)+c=7+c. Since p(1)=2, we have 7+c=2, so c=-5. If c were 5, then p(1) would be 12, not 2. Exam tip: In polynomial-value questions, substitute the given value of x first and then solve the resulting equation.

Given p(6)=0, substitute x=6 in p(x)=x+a. This gives p(6)=6+a, so 6+a=0 and hence a=-6. If a were 6, then p(6) would be 12, not 0. Exam tip: on substituting a zero of a polynomial, the polynomial value must be 0.

Given p(x)=3x-2, p(2)=3(2)-2=4 and p(-2)=3(-2)-2=-8. Hence, p(2)+p(-2)=4+(-8)=-4. The option 4 is only the value of p(2), not the required sum. Exam tip: when substituting a negative value, take care of the sign of the term 3x.

In the polynomial \(5x+5\), the coefficient of \(x\) is \(5\). To find its zero, set \(5x+5=0\). This gives \(5x=-5\), so \(x=-1\). Although \(5x-5\) also has coefficient \(5\), its zero is \(1\), not \(-1\). Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

A linear polynomial has the form \(ax+b\), where \(a\ne0\). Here \(a=-2\) and \(b=7\), so the degree is 1. A negative coefficient is valid; only \(a=0\) makes it constant. Exam tip: check the highest power, not the sign of the coefficient.

Given \(p(x)=5x+8\), we get \(p(x)-8=(5x+8)-8=5x\). The highest power of \(x\) in \(5x\) is \(1\), so its degree is \(1\). Option \(0\) would apply only if a non-zero constant term alone remained. Exam tip: simplify the expression first, then identify the highest exponent of the variable.

Given p(x)=3x-4, p(x)-3x=(3x-4)-3x=-4. Since -4 is a non-zero constant and contains no term in x, it is a constant polynomial. It is not a zero polynomial, because a zero polynomial must equal 0. Exam tip: simplify first, then identify the polynomial type from the degree of the remaining expression.

A zero of a polynomial is a value that makes the polynomial equal to 0. Substituting \(x=-\frac{4}{5}\) in \(5x+4\) gives \(5\left(-\frac{4}{5}\right)+4=-4+4=0\). Hence, option C is correct. Option A, \(5x-4\), has zero \(\frac{4}{5}\) because its constant term has the opposite sign. Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).

Given \(p(x)=12-5x\). Substituting \(x=2\), \(p(2)=12-5(2)=12-10=2\). Hence, the correct answer is 2. The number 5 is only the coefficient; the entire expression must be evaluated after substituting \(x=2\). Exam tip: When evaluating a polynomial, substitute the given value in brackets and perform multiplication before subtraction.

Given \(p(x)=2x-3\) and \(q(x)=x+5\), \(2p(x)-q(x)=2(2x-3)-(x+5)=4x-6-x-5=3x-11\). Hence, \(3x-11\) is correct. \(3x+11\) can result from incorrectly handling the minus sign before \((x+5)\). Exam tip: when subtracting a polynomial in brackets, change the sign of every term inside the bracket.

A zero of a polynomial is a value of \(x\) for which the polynomial equals 0. Put \(4x-8=0\): then \(4x=8\), so \(x=2\). Checking, \(p(2)=4(2)-8=0\); hence option B is correct. For example, at \(x=4\), the value is \(8\), not 0. Exam tip: To find a zero, set the polynomial equal to 0 and solve for \(x\).

The difference is found by subtracting the second polynomial from the first: \((3x+7)-(x-2)\). On removing the brackets, \(3x+7-x+2=2x+9\). In option B, the order of subtraction is reversed, so it gives a different result. Exam tip: while subtracting a bracket, change the sign of every term inside it.

In \(7x-5\), the highest power of \(x\) is 1, so it is a linear polynomial. \(x^2+3x-1\) is quadratic, while \(4/x\) and \(\sqrt{x}\) are not polynomials. Exam tip: powers must be non-negative integers.

Substituting \(x=12\), \(p(12)=\frac{5}{6}\times 12-10=10-10=0\). Hence, the correct answer is \(0\). The value \(-10\) is only the constant term, not the value of the polynomial at \(x=12\). Exam tip: For a polynomial with a fractional coefficient, multiply first and then add or subtract.

Here, (p(x)^2)=(x-2)^2=x^2-4x+4. The highest power of x is 2, so the polynomial has degree 2 and is a quadratic polynomial. A linear polynomial has degree 1, so option A is not correct. Exam tip: simplify the expression and check the highest power of the variable to identify the type of polynomial.

A linear polynomial has the form \(ax+b\), where \(a\ne 0\). In \(7x\), the coefficient of \(x\) is 7, which is non-zero; hence it is a linear polynomial with leading coefficient 7. \(12\) and \(-5\) are constant polynomials, while 0 is the zero polynomial; none has an \(x\)-term. Exam tip: To identify a linear polynomial, check that the highest power of \(x\) is 1 and its coefficient is non-zero.

Here, p(4)=3(4)+b=12+b and p(2)=3(2)+b=6+b. Therefore, p(4)-p(2)=(12+b)-(6+b)=6. The constant term b cancels from the difference. Option b is incorrect because b does not affect the difference. Exam tip: For a linear polynomial ax+b, p(m)-p(n)=a(m-n).

The degree of a polynomial is the highest power of \(x\) in any non-zero term. Since \(5x^2\) is present, its degree is 2, making it quadratic rather than linear. Exam tip: check the highest exponent after simplifying.

Since p(x) and q(x) are identical polynomials, p(x)-q(x)=(2x+3)-(2x+3)=0. This is the zero polynomial, whose degree is not defined. Do not treat \(0\) as a non-zero constant polynomial of degree 0. Exam tip: after subtracting polynomials, first check whether all terms cancel.

Given \(p(x)=9x-1\), we get \(p(1)=9(1)-1=8\) and \(p(0)=9(0)-1=-1\). Therefore, \(p(1)+p(0)=8+(-1)=7\). Note that \(8\) is only the value of \(p(1)\), not their sum. Exam tip: After substituting a value in a polynomial, check the signs of all terms carefully.

p(3)=5×3−4=11 and p(−1)=5×(−1)−4=−9. Therefore, p(3)−p(−1)=11−(−9)=20. The value 15 is only 5×3; it does not account for the constant term −4 or the value of p(−1). Exam tip: Use brackets when subtracting a negative number, since −(−9)=+9.

The governing concept is the degree of a polynomial. An expression in x is linear when the coefficient of x is nonzero, because then its degree is 1. In (2r+5)x−10, the coefficient of x is 2r+5. To make the expression no longer linear, set that coefficient equal to zero: 2r+5=0, so 2r=−5 and r=−5/2. Substitution gives 0x−10=−10, a constant polynomial of degree 0. Therefore option B is correct. For r=0, the coefficient is 5; for r=5, it is 15; and for r=5/2, it is 10, so each of those cases remains linear. The constant term −10 does not affect whether the x-term has degree one.

A zero of \(-2\) means that \(p(-2)=0\). Therefore, \(7(-2)+b=0\), so \(-14+b=0\). Hence, \(b=14\). If \(b=-14\), then \(p(-2)=-28\), not zero. Exam tip: For a linear polynomial \(ax+b\), substitute the given zero for \(x\) and set the polynomial equal to 0.

Given p(x)=2x-5 and q(x)=4x+1, 3p(x)-q(x)=3(2x-5)-(4x+1)=6x-15-4x-1=2x-16. Hence, (2x-16) is correct. In (2x+16), the signs of the terms in q(x) have not been changed correctly while subtracting it. Exam tip: When a minus sign occurs before a bracket, change the sign of every term inside the bracket.

A zero of a polynomial is a value of x that makes the polynomial equal to 0. For \(3x-7=0\), we get \(3x=7\), so \(x=\frac{7}{3}\). Hence, \(3x-7\) is correct. The close distractor \(3x+7\) has zero \(-\frac{7}{3}\), not \(\frac{7}{3}\). Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).