Linear polynomials

A zero is the value of x for which the polynomial becomes 0. Setting \(7x+21=0\) gives \(7x=-21\), so \(x=-3\). If \(x=3\), then \(7(3)+21=42\), so it is not a zero. Exam tip: the zero of a linear polynomial \(ax+b\) is \(-\frac{b}{a}\).

Given \(p(x)=5x-4\), we get \(p(4)=5\times4-4=16\) and \(p(2)=5\times2-4=6\). Hence, \(p(4)-p(2)=16-6=10\), so option D is correct. Option B can result from handling the constant term \(-4\) incorrectly; evaluate the polynomial completely at both inputs before subtracting. Exam tip: find each function value separately, then take their difference.

A zero of a polynomial is a value that makes the polynomial equal to 0. Hence, putting \(p(5)=0\) gives \(5m-15=0\). Therefore, \(5m=15\), so \(m=3\). If \(m=-3\), then \(5m-15=-30\), not 0. Exam tip: Substitute the given zero for \(x\) and set the polynomial equal to 0.

To find the zero of the linear polynomial \(ax+b\), set \(ax+b=0\). Then \(ax=-b\), so \(x=-\frac{b}{a}\), where \(a\ne0\). The option \(\frac{b}{a}\) misses the negative sign, so it is incorrect. Exam tip: remember that the zero of \(ax+b\) is \(-\frac{b}{a}\).

Adding the polynomials gives
(p(x)+q(x))=(6x+2)+(x-9)=7x-7. The highest power of x is 1, so its degree is 1. The number 7 is a coefficient, not the degree. Exam tip: First simplify by combining like terms, then identify the highest exponent of the variable.

p(x)-q(x)=(3x+10)-(3x-4)=3x+10-3x+4=14. The x-terms cancel, so the result, 14, is a non-zero constant polynomial. It is not a zero polynomial because its value is not 0. Exam tip: When subtracting a polynomial, change the sign of every term in the second polynomial.

If 3 is a zero of the polynomial, its value must be 0 when \(x=3\). Thus, \((k+2)\times 3-12=0\), or \(3k+6-12=0\). Hence \(3k=6\), so \(k=2\). For \(k=4\), the polynomial gives 6, not 0. Exam tip: Substitute the given zero into the polynomial and equate the result to 0.

In option B, combining like terms gives 9x-5x=4x. Hence, 9x-5x-7 simplifies to 4x-7. In option A, the coefficient of x becomes 12, while option C has the constant term +7, not -7. Exam tip: Add or subtract only terms with the same variable and exponent.

Substitute \(x=-3\) in the polynomial: \(p(-3)=4(-3)+3=-12+3=-9\). Hence, the correct value is \(-9\). The distractor \(-15\) can result from incorrectly treating the constant \(+3\) as negative. Exam tip: Put a negative substituted value in brackets before multiplying.

In \(5x-2\), the coefficient of \(x\) is \(5\), which is positive, and the constant term is \(-2\), which is negative. Therefore, option D is correct. In \(-5x-2\), the constant term is negative, but the coefficient of \(x\), \(-5\), is also negative. Exam tip: in a linear polynomial \(ax+b\), \(a\) is the coefficient of \(x\) and \(b\) is the constant term.

Given \(p(x)=ax+18\), substituting \(x=-3\) gives \(p(-3)=-3a+18\). Since \(p(-3)=0\), we get \(-3a+18=0\), so \(a=6\). If \(a=-6\), then \(p(-3)=36\), not zero. Exam tip: Substitute the given value of \(x\) carefully before solving for the unknown coefficient.

On substituting x=-2, we get x+2=(-2)+2=0. Hence, -2 is a zero of x+2, so option C is correct. For example, substituting -2 in x-2 gives -4, not 0. Exam tip: To check whether a number is a zero of a polynomial, substitute it and see whether the value obtained is 0.

On opening the brackets, \(5(x-1)+3(x+2)=5x-5+3x+6\). Combining like terms gives \(5x+3x=8x\) and \(-5+6=1\), so the polynomial is \(8x+1\). \(8x-1\) is the closest distractor, but it uses an incorrect sum of the constant terms. Exam tip: Multiply the number outside each bracket by every term inside it.

To find the zero, set \(p(x)=0\): \(14-7x=0\). Thus, \(7x=14\), so \(x=2\). Substituting \(-2\) gives \(14-7(-2)=28\), not zero. Exam tip: verify a zero by substituting it into the polynomial; the result must be \(0\).

A linear polynomial has the form \(ax+b\), where \(a\ne0\). For \(p(x)=2x+8\), \(p(0)=2(0)+8=8\), so option B is correct. In option C, \(p(0)=-8\), while option D has degree 0 and is a constant polynomial, not a linear polynomial. Exam tip: To find \(p(0)\), substitute 0 for \(x\); the result is the constant term.

Given p(x)=6x+1, substituting t for x gives p(t)=6t+1. Since p(t)=31, we get 6t+1=31, so 6t=30 and t=5. For example, t=4 gives p(4)=25, not 31. Exam tip: In p(t), replace x directly with t before solving the equation.

To find the zero of the linear polynomial \(x-6\), set \(x-6=0\). This gives \(x=6\), which is positive. In contrast, the zero of \(x+7\) is \(-7\). Exam tip: The zero of \(ax+b\) is \(-\frac{b}{a}\); check its sign carefully.

A zero of \(-5\) means that \(p(-5)=0\). Thus, \(4(-5)+a=0\), so \(-20+a=0\). Therefore, \(a=20\). If \(a=-20\), then \(p(-5)=-40\), not 0. Exam tip: Substitute the given zero for \(x\) and set the polynomial equal to 0.

A zero of a linear polynomial is the value of x for which the polynomial becomes 0. Substituting \(x=\frac{4}{3}\) in \(3x-4\) gives \(3\times\frac{4}{3}-4=0\). Hence, \(3x-4\) has zero \(\frac{4}{3}\). The close distractor \(4x-3\) has zero \(\frac{3}{4}\), not \(\frac{4}{3}\). Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).

\(p(x)q(x)=(x+4)(x-4)=x^2-16\), using the identity \((a+b)(a-b)=a^2-b^2\). 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 option A is not correct. Exam tip: Before multiplying, look for conjugate factors to apply the difference-of-squares identity.