Linear polynomials

To find a zero, set the polynomial equal to 0: \(5x+20=0\). Thus, \(5x=-20\), so \(x=-4\). Substituting \(x=4\) gives \(5(4)+20=40\), not 0. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

Given \(p(x)=7x-9\), \(p(2)=7\times2-9=5\) and \(p(1)=7\times1-9=-2\). Hence, \(p(2)-p(1)=5-(-2)=7\). Option 9 is only the constant term’s number, not the required difference. Exam tip: After substituting values in a polynomial, handle subtraction of a negative number carefully.

In option A, \(x^2-x^2=0\), so the expression simplifies to \(x+4\). Its degree is 1, so it is a linear polynomial. Option B has degree 2 and is therefore quadratic, while \(7\) is a constant polynomial. In exams, first combine like terms and then identify the highest power of the variable.

A zero of a polynomial is a value of x for which the polynomial equals 0. Substituting x=2 gives p(2)=2m+6=0. Hence, 2m=-6 and m=-3. The option -6 results if one forgets to divide by 2. Exam tip: when a zero is given, substitute it for x and set p(x)=0.

To find the zero of a linear polynomial, set \(ax+b=0\). Then \(ax=-b\), so \(x=-\frac{b}{a}\), where \(a\ne0\) is necessary. The expression \(\frac{b}{a}\) misses the negative sign, so it is not the correct zero. Exam tip: Substitute the obtained value in \(ax+b\); it must give 0.

Here, p(x)+q(x)=(2x-3)+(4x+5)=6x+2. The highest power of x in this polynomial is 1, so its degree is 1. Option 6 is the coefficient of x, not the degree. Exam tip: simplify the polynomial first, then identify the highest exponent of the variable.

On subtracting, \(p(x)-q(x)=(4x+1)-(4x-7)=4x+1-4x+7=8\). The \(x\)-terms cancel, leaving the non-zero constant 8, so the result is a constant polynomial. A linear polynomial must have a non-zero coefficient of \(x\), which is absent here. Exam tip: While subtracting polynomials, change the sign of every term in the second bracket.

If \(x=1\) is a zero of \(((k+5)x-11)\), then \((k+5)(1)-11=0\). Thus, \(k+5-11=0\), so \(k=6\). For \(k=-5\), the coefficient of \(x\) becomes zero and the expression is \(-11\), which has no zero. Exam tip: substitute the given zero into the polynomial and equate the result to zero.

In option B, combining like terms gives \(8x-5x-2=(8-5)x-2=3x-2\). In option A, the constant term is \(+2\), so it simplifies to \(3x+2\), not \(3x-2\). Exam tip: First combine the coefficients of like \(x\)-terms, then check the sign of the constant term carefully.

Substituting -2 for x gives \(p(-2)=3(-2)+2=-6+2=-4\). Therefore, the correct answer is -4. The value -8 results from incorrectly treating the constant term +2 as a subtraction. In exams, multiply the negative number first and then perform addition or subtraction.

In \(-4x-3\), the coefficient of \(x\) is \(-4\) and the constant term is \(-3\). Hence, both are negative. In option B, the coefficient of \(x\) is negative, but the constant term \(+3\) is positive. Exam tip: In a linear polynomial \(ax+b\), \(a\) is the coefficient of \(x\) and \(b\) is the constant term.

Given p(5)=0, substitute x=5 in the polynomial: 5a-10=0. Thus, 5a=10 and a=2. The value 5 would be obtained only if 5a=25, which is not the case here. Exam tip: while evaluating a polynomial, substitute the given value for every occurrence of x carefully.

On substituting \(x=3\), we get \(x-3=3-3=0\). Hence, 3 is a zero of \(x-3\), so it is the correct linear polynomial. In contrast, \(x+3\) gives 6 at \(x=3\), not zero. Exam tip: To check whether a number is a zero of a polynomial, substitute it and verify that the value is 0.

Using the distributive property, \(2(x-4)=2x-8\). Therefore, \(2x-8+5x=(2x+5x)-8=7x-8\). The option \(7x+8\) results from incorrectly handling the negative sign when multiplying 2 by \(-4\). Exam tip: when opening brackets, multiply the outside number by every term inside the bracket.

A zero of a polynomial is a value of \(x\) for which the polynomial becomes 0. Here, \(6-2x=0\) gives \(2x=6\), so \(x=3\). Therefore, 3 is the correct answer. If \(x=2\), then \(p(2)=2\), so it is not a zero. Exam tip: To find a zero, set \(p(x)=0\) and solve for \(x\).

To find \(p(0)\), substitute \(x=0\). For \(p(x)=x+5\), \(p(0)=0+5=5\), and its degree is 1, so it is a linear polynomial. Although \(p(x)=5\) also gives \(p(0)=5\), it is a constant polynomial, not a linear one. Exam tip: a linear polynomial has highest power 1.

Given \(p(x)=5x-2\), we get \(p(t)=5t-2\). Substituting \(p(t)=18\) gives \(5t-2=18\), so \(5t=20\) and hence \(t=4\). For example, \(p(3)=13\), not 18. Exam tip: To find \(p(t)\), directly replace \(x\) with \(t\).

In option C, the polynomial is \(x+4\). From \(x+4=0\), its zero is \(x=-4\). Since \(-4\) is not positive, C is correct. In contrast, the zero of \(x-5\) is \(5\), which is positive. Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).

A zero of a polynomial is a value for which the polynomial becomes 0. Hence, \(p(-3)=0\). So, \(2(-3)+a=0\Rightarrow -6+a=0\Rightarrow a=6\). If \(a=-6\), then \(2(-3)-6=-12\), not 0. Exam tip: Substitute the given zero for \(x\) and set the polynomial equal to 0.

A zero of a linear polynomial is a value of x that makes the polynomial equal to 0. On substituting \(x=\frac{3}{2}\), \(2x-3=2\left(\frac{3}{2}\right)-3=0\); hence option B is correct. Option C, \(3x-2\), has zero \(\frac{2}{3}\), not \(\frac{3}{2}\). Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).