Question

type the correct answer in the box. use numerals instead of words. if necessary, use / for the fraction bar. consider the expression \\(\frac{x^2 + x - 10}{x + 3}\\). when using the inspection method, the number you would add to (and subtract from) the constant term of the numerator so the polynomial in the numerator will have (x + 3) as a factor is \\(\boxed{}\\). reset next

Explanation:

Step1: Recall the factor theorem

If \((x + 3)\) is a factor of \(f(x)=x^{2}+x - 10\), then \(f(-3)=0\). Let's find \(f(-3)\): \(f(-3)=(-3)^{2}+(-3)-10=9 - 3-10=-4\). But we can also use the inspection method. We want to write \(x^{2}+x - 10\) as \(x^{2}+x + c - c-10\) such that \(x^{2}+x + c\) has \((x + 3)\) as a factor. If \((x + 3)\) is a factor of \(x^{2}+x + c\), then when we substitute \(x=-3\) into \(x^{2}+x + c\), we get \(0\). So \((-3)^{2}+(-3)+c = 0\), which simplifies to \(9-3 + c=0\), so \(c=-6\). But we need to find the number to add and subtract from the constant term. The original constant term is \(-10\). Let the number be \(k\). So we have \(x^{2}+x-10 + k - k=x^{2}+x+(k - 10)-k\). We want \(x^{2}+x+(k - 10)\) to have \((x + 3)\) as a factor. Using the factor theorem, substitute \(x = - 3\) into \(x^{2}+x+(k - 10)\): \((-3)^{2}+(-3)+(k - 10)=0\). Calculate: \(9-3 + k-10 = 0\), \(k-4 = 0\), \(k = 4\)? Wait, no, wait. Wait, the standard method for inspection (completing the factor) for dividing by \((x + a)\) is to make the numerator have \((x + a)\) as a factor. Let's do polynomial division or use the fact that if we have \(x^{2}+x - 10\), and we want to factor out \((x + 3)\), we can write \(x^{2}+x - 10=(x + 3)(x + b)+d\). Expand the right side: \(x^{2}+(b + 3)x+3b + d\). Equate coefficients: \(b + 3 = 1\) (coefficient of \(x\)), so \(b=-2\). Then constant term: \(3b + d=-10\). Substitute \(b = - 2\): \(3(-2)+d=-10\), \(-6 + d=-10\), \(d=-4\). But we want to adjust the constant term. Alternatively, let's find the value that when added to the constant term (and subtracted) makes the quadratic factorable with \((x + 3)\). Let the numerator be \(x^{2}+x-10 + k - k\). We want \(x^{2}+x+(k - 10)\) to be divisible by \((x + 3)\). So when \(x=-3\), \(x^{2}+x+(k - 10)=0\). So \((-3)^2+(-3)+(k - 10)=0\) → \(9 - 3 + k - 10 = 0\) → \(k - 4 = 0\) → \(k = 4\)? Wait, no, that's not right. Wait, maybe I messed up. Let's try another approach. The general form: if we have \(\frac{x^{2}+x - 10}{x + 3}\), we can write the numerator as \(x^{2}+3x-2x-10\) (but that's factoring by grouping). Wait, the inspection method for adding and subtracting a number to the constant term. Let's let the number be \(k\). So we have \(x^{2}+x-10 + k - k=x^{2}+x+(k - 10)-k\). We want \(x^{2}+x+(k - 10)\) to have \((x + 3)\) as a factor. So the roots of \(x^{2}+x+(k - 10)\) should include \(x=-3\). So substitute \(x=-3\) into \(x^{2}+x+(k - 10)\): \(9-3 + k - 10 = 0\) → \(k - 4 = 0\) → \(k = 4\)? Wait, no, that gives \(x^{2}+x+(4 - 10)=x^{2}+x - 6\), which factors as \((x + 3)(x - 2)\). Ah! Yes, \(x^{2}+x - 6=(x + 3)(x - 2)\). So we added \(4\) to \(-10\) (because \(-10+4=-6\)) and then subtracted \(4\). So the number we add (and subtract) from the constant term is \(4\)? Wait, no, the constant term was \(-10\), we added \(4\) to get \(-6\), then subtracted \(4\). Wait, but let's check: \(x^{2}+x-10=x^{2}+x-6 - 4=(x + 3)(x - 2)-4\). So when we use the inspection method, we want to make the numerator have \((x + 3)\) as a factor, so we need to adjust the constant term so that the quadratic is divisible by \((x + 3)\). The quadratic \(x^{2}+x + c\) divisible by \((x + 3)\) must satisfy that when \(x=-3\), \(x^{2}+x + c = 0\), so \(9-3 + c=0\) → \(c=-6\). The original constant term is \(-10\), so we need to add \(4\) to \(-10\) to get \(-6\) (since \(-10 + 4=-6\)), and then subtract \(4\) to keep the expression the same. So the number is \(4\)? Wait, no, wait: the constant term of the numerator is \(-10\). We need to add \(k\) and subtract \(k\), so the numerator becomes \(x^{2}+x+(-10 + k)-k\). We want \(x^{2}+x+(-10 + k)\) to have \((x + 3)\) as a factor. So set \(x=-3\) in \(x^{2}+x+(-10 + k)\): \((-3)^2+(-3)+(-10 + k)=0\) → \(9-3-10 + k=0\) → \(-4 + k=0\) → \(k = 4\). Yes, that's correct. So the number to add (and subtract) from the constant term is \(4\)? Wait, but let's check with the quadratic \(x^{2}+x-10 + 4=x^{2}+x-6\), which factors as \((x + 3)(x - 2)\), which is divisible by \((x + 3)\). Then we subtract \(4\) to get back to the original expression: \(x^{2}+x-6-4=x^{2}+x-10\). So yes, the number is \(4\)? Wait, no, wait, I think I made a mistake earlier. Wait, the formula: when dividing \(ax^{2}+bx + c\) by \((x + m)\), the inspection method involves writing \(ax^{2}+bx + c=ax^{2}+bx+am^{2}-am^{2}+c\) (no, better to use the factor theorem). Wait, the correct way: Let \(f(x)=x^{2}+x-10\). We want to write \(f(x)=(x + 3)q(x)+r\), but we want to make \(r = 0\) by adjusting the constant term. Wait, no, the question is: "the number you would add to (and subtract from) the constant term of the numerator so the polynomial in the numerator will have \((x + 3)\) as a factor". So the numerator is \(x^{2}+x-10\). Let the number be \(k\). So we have \(x^{2}+x-10 + k - k=x^{2}+x+(k - 10)-k\). We want \(x^{2}+x+(k - 10)\) to have \((x + 3)\) as a factor. So, by factor theorem, \(f(-3)=0\) for \(f(x)=x^{2}+x+(k - 10)\). So \((-3)^2+(-3)+(k - 10)=0\) → \(9-3 + k - 10=0\) → \(k - 4=0\) → \(k = 4\). So the number is \(4\)? Wait, but let's check with \(k = 4\): numerator becomes \(x^{2}+x-10 + 4-4=x^{2}+x-6-4\). Wait, no, \(x^{2}+x-10 + 4=x^{2}+x-6\), which has \((x + 3)\) as a factor (since \(x^{2}+x-6=(x + 3)(x - 2)\)), and then we subtract \(4\) to get back to the original numerator. So the number to add (and subtract) is \(4\)? Wait, but I think I messed up the sign. Wait, the constant term is \(-10\). We need to add \(k\) to \(-10\) to get a constant term \(c\) such that \(x^{2}+x + c\) is divisible by \((x + 3)\). As we found, \(c=-6\), so \(k=c-(-10)=-6 + 10 = 4\). Yes, so \(k = 4\). So the number is \(4\).

Step2: Verify

Let's check the numerator after adding and subtracting \(4\): \(x^{2}+x-10+4 - 4=x^{2}+x-6-4\). The polynomial \(x^{2}+x-6\) factors as \((x + 3)(x - 2)\), which has \((x + 3)\) as a factor. So the number we add (and subtract) from the constant term is \(4\).

Answer:

4