Linear polynomials

Given \(p(x)=6x-5\), \(p(4)=6\times4-5=19\) and \(p(1)=6\times1-5=1\). Therefore, \(p(4)-p(1)=19-1=18\). The value 24 comes from taking only \(6\times4\), which ignores the constant term and the subtraction of \(p(1)\). Exam tip: evaluate the polynomial separately at each given value before finding their difference.

For a polynomial to be linear, the coefficient of \(x\) must not be zero. Here, the coefficient of \(x\) is \(3r-12\). For the expression to stop being linear, \(3r-12=0\), which gives \(r=4\). At this value, the expression becomes \(5\), a constant polynomial. For the nearby option \(r=3\), the coefficient is \(-3\), so it is still linear. Exam tip: In a parameter-based polynomial, set the coefficient of the highest power to zero to check when its degree decreases.

A zero of a polynomial is a value of x for which the polynomial becomes 0. Substituting x=-1 gives 9(-1)+b=0. Thus, -9+b=0, so b=9. If b=-9, then p(-1)=-18, not 0. Exam tip: When a zero is given, substitute it in the polynomial and equate the result to 0.

First multiply \(q(x)\) by 2: \(2q(x)=2(5x+4)=10x+8\). Now subtract \(p(x)\): \(10x+8-(3x-2)=10x+8-3x+2=7x+10\). Hence, \(7x+10\) is correct. \(7x+6\) results from mishandling the sign of \(-2\) while subtracting \((3x-2)\). Exam tip: when a minus sign appears before a polynomial, change the signs of all terms inside the bracket.

A zero of a linear polynomial is the value of x that makes the polynomial equal to 0. From \(5x+2=0\), we get \(5x=-2\), so \(x=-\frac{2}{5}\). Therefore, option C is correct. Option A has zero \(\frac{2}{5}\) because its constant term is \(-2\). Exam tip: the zero of \(ax+b\) is \(-\frac{b}{a}\).

Substituting \(x=-3\), we get \(p(-3)=7-2(-3)=7+6=13\). Therefore, the correct value is \(13\). The option \(-13\) may result from handling the signs incorrectly. Exam tip: always substitute a negative value using brackets.

In option A, both polynomials are identical, so their difference is \((2x+3)-(2x+3)=0\). Hence, the result is the zero polynomial. In option B, the difference is \(6\), not zero. Exam tip: For the difference of two polynomials to be zero, the coefficients of their corresponding terms must be equal.

Substituting \(x=10\), \(p(10)=\frac{4}{5}\times10-8=8-8=0\). Hence, the correct answer is 0. \(-8\) is only the constant term, not the value of \(p(10)\). Exam tip: While evaluating a polynomial, substitute the given value of \(x\) carefully in every term.

On opening the brackets, \(5(2x-3)=10x-15\) and \(-3(3x+1)=-9x-3\). Therefore, \(10x-15-9x-3=x-18\). Hence, the correct polynomial is \(x-18\). In \(10x-18\), the \(-9x\) term has not been combined. Exam tip: When opening a bracket preceded by a negative sign, change the sign of every term inside it.

For a polynomial to be linear in x, the coefficient of x must never be zero. In option A, \(p^2+1>0\) for every real p, so it always has degree 1. In option B, the coefficient becomes zero at \(p=0\) or \(1\). Exam tip: always test whether a parameter-dependent coefficient can vanish.

A zero of a polynomial is a value for which the polynomial becomes 0. So, put \(p(-3)=0\): \((k-2)(-3)+9=0\). This gives \(-3k+6+9=0\), or \(-3k+15=0\), hence \(k=5\). If \(k=2\), the polynomial reduces to the constant 9, so it cannot have \(-3\) as a zero. Exam tip: When a zero is given, substitute it for \(x\) and equate the polynomial to 0.

\(2p(x)-3q(x)=2(3x-2)-3(x+5)\). On expanding, \(6x-4-3x-15=3x-19\). Hence, the correct expression is \(3x-19\). The option \(x-19\) results from incorrectly simplifying \(6x-3x\) as \(x\). Exam tip: When a minus sign occurs before brackets, apply it to every term inside the bracket.

If \(x=4\) is a zero, substituting \(x=4\) in \((2m-5)x+10\) must give zero. Thus, \(4(2m-5)+10=0\). On simplifying, \(8m-20+10=0\), so \(8m=10\) and hence \(m=\frac{5}{4}\). For \(m=\frac{5}{2}\), the coefficient of \(x\) becomes zero and the expression is \(10\), which has no zero. Exam tip: Substitute the given zero in the polynomial and equate the result to zero.

A zero of 2 means \(p(2)=0\). Therefore, \(2a+b=0\). Substituting \(a=5\) gives \(2\times5+b=0\), so \(b=-10\). If \(b=10\), then \(p(2)=20\), not zero. Exam tip: for a zero \(r\) of a linear polynomial \(ax+b\), write \(ar+b=0\).

A linear polynomial must have highest power 1, so the coefficient \(a\) of \(x\) must be non-zero. \(b\) may be zero; \(b\ne0\) alone does not ensure linearity. Exam tip: check the coefficient of \(x\) first.

For \(p(x)=4x+c\), we have \(p(3)=12+c\) and \(p(1)=4+c\). Hence, \(p(3)-p(1)=(12+c)-(4+c)=8\), but the given condition requires this difference to be 10. The constant \(c\) cancels out, so no value of \(c\) can satisfy the condition. Option B is incorrect because the difference remains 8 for every \(c\). Exam tip: When comparing two values of a linear polynomial, the constant term often cancels out.

A zero of a polynomial is the value of the variable that makes the polynomial equal to zero. For a linear polynomial, we find it by equating the expression to zero and solving for the variable. This is different from merely looking at the constant term; both terms must be considered together.

For option C, set \(7x+3=0\). Subtracting 3 gives \(7x=-3\), and dividing by 7 gives \(x=-\frac{3}{7}\). Therefore option C has the required zero. Option A would give \(x=\frac{3}{7}\), while option B gives \(-\frac{7}{3}\); hence neither has the stated value.

The governing concept is the zero of a linear polynomial. A zero is the value of x for which the polynomial evaluates to 0. Set p(x) equal to zero: (2/5)x − 6 = 0. Adding 6 to both sides gives (2/5)x = 6. Multiply both sides by the reciprocal 5/2: x = 6 × 5/2 = 15. Substitution verifies the result because p(15) = (2/5)(15) − 6 = 6 − 6 = 0. Therefore, option C is correct. The other values do not make the expression zero and can arise from incorrectly handling the fractional coefficient, especially by multiplying or dividing by 5 without also treating the numerator correctly.

Both given polynomials are linear, so each has degree 1. On multiplying,
\((x+2)(x-5)=x^2-3x-10\). The highest power of the variable is 2, so the degree of the product is 2. Option 1 is the degree of each individual linear polynomial, not of their product. Exam tip: for non-zero polynomials, the degree of a product is generally the sum of their degrees.

Replace the entire variable \(x\) in the polynomial with \(x+3\). Thus, \(p(x+3)=2(x+3)-1=2x+6-1=2x+5\). Therefore, \(2x+5\) is correct. The option \(x+5\) incorrectly drops the coefficient \(2\). Exam tip: always use brackets when substituting an expression for a variable.