Linear polynomials

Substituting \(x=4\), we get \(p(4)=\frac{1}{2}\times 4+3=2+3=5\). Hence, 5 is the correct option. Option 4 may result from incorrectly evaluating \(\frac{1}{2}\times4\) or from not adding 3. Exam tip: To find the value of a polynomial, substitute the given number for the variable, then perform multiplication before addition or subtraction.

For \((2a+1)x-4\) to be a linear polynomial, the coefficient of \(x\) must not be zero. It will cease to be linear when \(2a+1=0\). Hence, \(a=-\frac{1}{2}\), and the expression becomes \(-4\), a constant polynomial rather than a linear polynomial. For example, at \(a=0\), the coefficient of \(x\) is 1, so it is still linear. Exam tip: In a parameter-based linear polynomial, set the coefficient of the variable equal to zero to find when it is not linear.

Given \(p(x)=10x-15\) and \(p(x)=5\), we get \(10x-15=5\). Adding 15 to both sides gives \(10x=20\), so \(x=2\). If \(x=1\), then \(p(1)=-5\), not 5. Exam tip: To find the input for a given polynomial value, equate \(p(x)\) to that value and solve the resulting equation.

For a linear polynomial \(ax+b\), putting \(x=0\) gives \(b\), the constant term. For \(2x-7\), \(2(0)-7=-7\). Although \(7x-2\) has a negative constant term, its value at zero is \(-2\), not \(-7\). Exam tip: At \(x=0\), identify the polynomial’s value directly from its constant term.

Given p(x)=4x+c, substitute x=2: p(2)=4(2)+c=8+c. Since p(2)=11, we get 8+c=11, so c=3. Option 4 could result from incorrectly evaluating 4x. Exam tip: substitute the given value of x carefully before solving for the unknown constant.

In \(0x+5\), \(0x=0\), so the polynomial simplifies to \(5\). Since \(5\) is a non-zero constant polynomial, its degree is 0; therefore, stating its degree as 1 is incorrect. In the other three polynomials, the coefficient of \(x\) is non-zero, so each has degree 1. Exam tip: simplify a polynomial by removing zero-coefficient terms before finding its degree.

Given \(p(x)=x+a\) and \(p(-5)=0\), substituting \(x=-5\) gives \(-5+a=0\). Hence, \(a=5\). If \(a=-5\), then \(p(-5)=-10\), not zero. Exam tip: when a zero of a polynomial is given, substitute that value of \(x\) and solve the resulting equation.

\(5x\) is a linear polynomial because the highest power of \(x\) is 1. To find its zero, set \(5x=0\); this gives \(x=0\). Although \(x^2\) also has zero 0, it is a quadratic polynomial because its degree is 2, not linear. Exam tip: a linear polynomial always has degree 1.

Given p(x)=2x-1, p(3)=2(3)-1=5 and p(-1)=2(-1)-1=-3. Hence, p(3)+p(-1)=5+(-3)=2. Option 1 may result from an incorrect operation with the negative value. Exam tip: Substitute each input separately and find p(3) and p(-1) before adding them.

In the polynomial 3x+6, the coefficient of x is 3. To find its zero, set 3x+6=0: 3x=-6, so x=-2. Therefore, 3x+6 is the correct option. The zero of 3x-6 is 2, not -2. Exam tip: the zero of ax+b is -b/a.

In \(5-3x\), the highest power of \(x\) is 1 and the coefficient \(-3\neq0\), so it is a linear polynomial. A negative coefficient is allowed. Exam tip: check the degree, not the sign of the coefficient.

In option B, \((3x+4)+(-3x+5)=3x-3x+4+5=9\). The \(x\)-terms cancel each other, so the sum is the constant polynomial \(9\). In option C, the constant terms cancel, but the sum is \(2x\), which is a linear polynomial. Exam tip: for the sum to be a constant, the coefficients of \(x\) in the two polynomials must be opposites.

Given \(p(x)=3x+4\), we get \(p(x)-4=(3x+4)-4=3x\). The highest power of \(x\) in \(3x\) is 1, so its degree is 1. Option 0 would apply only if the result were a non-zero constant. Exam tip: simplify the expression first, then identify the highest exponent of the variable.

Given p(x)=2x+7, p(x)-2x=(2x+7)-2x=7. There is no term containing x, and the value is non-zero, so it is a non-zero constant polynomial. A zero polynomial is only 0, whereas the resulting polynomial here is 7. Exam tip: After simplifying, identify the polynomial type from the highest power of x present.

A zero of a polynomial is a value that makes the polynomial equal to 0. Substituting \(x=-\frac{5}{3}\) in \(3x+5\) gives \(3\left(-\frac{5}{3}\right)+5=-5+5=0\). Hence, option C is correct. Option A has zero \(\frac{5}{3}\), since its constant term has a negative sign. Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).

Given \(p(x)=9-4x\). Substituting \(x=2\), we get \(p(2)=9-4(2)=9-8=1\). Hence, the correct answer is 1. The value 5 would be obtained for \(x=1\), not for \(x=2\). Exam tip: To find the value of a polynomial, substitute the given value for every \(x\), then perform multiplication before addition or subtraction.

In option B, \(7x-7x+3=(7-7)x+3=3\). Therefore, the coefficient of \(x\) is \(0\). In option A the coefficient is \(2\), and in option C it is \(3\). Exam tip: while combining like terms, add or subtract the coefficients of the \(x\)-terms.

Given \(p(x)=x-4\) and \(q(x)=2x+1\), \(2p(x)+q(x)=2(x-4)+(2x+1)=2x-8+2x+1=4x-7\). Hence, \(4x-7\) is correct. The option \(3x-7\) may result from incorrectly combining the \(x\)-terms in \(2p(x)\). Exam tip: Multiply each polynomial by its coefficient first, then combine like terms.

A zero of a polynomial is a value of \(x\) for which the polynomial equals 0. Setting \(p(x)=3x-6=0\) gives \(3x=6\), so \(x=2\). For example, substituting \(x=1\) gives \(-3\), so it is not a zero. Exam tip: To find a zero, always solve \(p(x)=0\).

To find the difference, subtract the second expression from the first: \((2x+5)-(x-1)\). On removing the bracket, \(2x+5-x+1=x+6\). In option B, the order of subtraction is reversed, so it gives \(-x-6\). Exam tip: when removing brackets preceded by a minus sign, change every sign inside the bracket.

Riya is incorrect. The degree depends on the highest exponent of the variable: in \(5-2x=-2x+5\), the exponent of \(x\) is 1, so it is linear. A negative coefficient does not change the degree. Exam tip: check the highest power, not the number of terms.

Substituting \(x=8\), \(p(8)=\frac{3}{4}\times 8-6=6-6=0\). Hence, the correct answer is 0. Option 6 is only the value of \(\frac{3}{4}\times8\); subtracting 6 is also necessary. Exam tip: While evaluating a polynomial, substitute the given value of \(x\) carefully in every term.

Here, \(p(x)^2=(x+3)^2=x^2+6x+9\). The highest power of \(x\) is 2, so its degree is 2 and it is a quadratic polynomial. A linear polynomial has degree 1, so that option is incorrect. Exam tip: Simplify the expression first, then identify the highest power of the variable.

In \(5x+0=5x\), the coefficient of \(x\) is \(5\), which is non-zero. Therefore, its degree is 1 and it is a linear polynomial. \(0x+9\) and \(0x-3\) simplify to \(9\) and \(-3\), so they are constant polynomials; \(7\) is also constant. Exam tip: for a linear polynomial, the coefficient of the variable must be non-zero.

Here, p(3)=2(3)+b=6+b and p(1)=2(1)+b=2+b. Thus, p(3)-p(1)=(6+b)-(2+b)=4, whereas the given condition requires the difference to be 8. The value of b cancels out, so no value of b can satisfy the condition. Exam tip: In such questions, calculate both function values first and then compare their difference.