32Number 32ExplanationIf h(x)h(x)h(x) is the remainder of dividing f(x)=5x4−2x2+7x+9f(x) = 5x^4 - 2x^2 + 7x + 9f(x)=5x4−2x2+7x+9 by x2−5x^2 - 5x2−5. The value of h(1)h(1)h(1) is....121121121131131131141141141151151151161161161ExplanationIt's easy, just use the long division method or "porogapit" as follows: 5x2+23x2−55x4+0x3−2x2+7x+95x4+0x3−25x2+7x+9‾ −5x4+0x3 23x2+7x+95x4+0x3 23x2+0x−115‾ −5x4+0x3−2x2 7x+124\begin{array}{c|l} & 5x^2 + 23 \\ \hline x^2 - 5 & 5x^4 + 0x^3 - 2x^2 + 7x + 9 \\ & \underline{5x^4 + 0x^3 - 25x^2 \phantom{+ 7x + 9}} \; - \\ & \phantom{5x^4 + 0x^3} \; 23x^2 + 7x + 9 \\ & \phantom{5x^4 + 0x^3} \; \underline{23x^2 + 0x - 115} \; - \\ & \phantom{5x^4 + 0x^3 - 2x^2} \; 7x + 124 \end{array}x2−55x2+235x4+0x3−2x2+7x+95x4+0x3−25x2+7x+9−5x4+0x323x2+7x+95x4+0x323x2+0x−115−5x4+0x3−2x27x+124 The remainder of the division is h(x)h(x)h(x). Based on the method above, the remainder is 7x+1247x + 1247x+124. So the value of h(1)h(1)h(1) h(x)=7x+124h(x) = 7x + 124h(x)=7x+124h(1)=7(1)+124=131h(1) = 7(1) + 124 = 131h(1)=7(1)+124=131
