# Simplify 3x^0 Assessor’s use only QUESTION ONE (a) (3x + 2)(x – 4)2 Expand and simplify: (3x + 2)(x2 – 8x + 16) (x2 – 8x + 16) (x2 – 8x + 16) 3x × 2× (b) A 𝟏𝟎𝒙𝟐 − 𝒙 − 𝟐 Simplify: 𝟔𝒙 − 𝟑 10𝑥 2 − 𝑥 − 2 6𝑥 − 3 (c) = 3x3 – 24x2 + 48x = 2x2 – 16x + 32 2 = 3x – 22x2 + 32x + 32 (5𝑥 + 2)(2𝑥 − 1) = 3(2𝑥 – 1) = 𝟓𝒙 + 𝟐 𝟑 top line factorised A fully cancelled M Katya is going use a picture frame that is 15 cm by 20 cm to hold a picture. The width of the coloured edging around the outside of the picture (to the edge of the frame) is the same all the way round. The area of the coloured edging inside the frame is cm2. Calculate the width of the coloured edging. Area of strip = 20 × 15 – (20 – 2x)(15 – 2x) = or 20x + 15x + 20x + 15x – 4x2 = or = 2 × 20x + 2 × (15 – 2x)x etc Equation to solve is = 70x – 4x2 A 4x2 – 70x + = 0 Solves to x = 15 or 15 makes no sense, as the picture is only 15cm wide, so strip is cm wide (d) 𝟐 Simplify: 𝒙−𝟏 𝟐 𝒙−𝟏 = – – 𝟓 𝒙+𝟑 = M 𝟓 𝒙+𝟑 𝟐(𝒙 + 𝟑) (𝒙 – 𝟏)(𝒙 + 𝟑) + −𝟓(𝒙 − 𝟏) (𝒙 + 𝟑)(𝒙 − 𝟏) = 𝟐𝒙 + 𝟔 − 𝟓𝒙 + 𝟓 (𝒙 – 𝟏)(𝒙 + 𝟑) −𝟑𝒙 + 𝟏𝟏 A (𝒙 − 𝟏)(𝒙 + 𝟑) or numerator may be 11 – 3x or denominator may be x2 + 4x + 3 1 Assessor’s use only (e) A rectangle is twice as wide as it is high. The height is then increased by the half the amount the width is decreased. → The result is that the area of the second rectangle is smaller than that of the first rectangle. How much higher is the second rectangle than the first? Original rectangle is x × 2x. New rectangle is (x + k)(2x – 2k) or equivalent A Equation is 2x2 – (x + k)(2x – 2k) = 2x2 – (2x2 – 2kx + 2kx – 2k2) = 2x2 = k = ± M The question states that the second rectangle is taller, so only the positive value applies in this case. (Or can regard negative answer as negative “increase”) The second rectangle is higher E (Cannot get E without at least some indication of the negative answer) (f) A isosceles triangular support is 14 metres wide, and 7 metres high. A diagonal supporting beam is 10 metres long. How far off the ground is the high end of the beam? The right sloping line has the equation y = – x + 14 By Pythagoras, = x2 + y2 = x2 + (14 – x) 2 A Expanding, = x2 + – 28x + x2, which gives 2x2 – 28x + 96 = 0 Solving gives x = 6 or 8 M x = 6 is outside the triangle, so x = 8. Using y = –x + 14, the height is 6 m E 2 Assessor’s use only QUESTION TWO (a) Solve: 4x2 – 9 = 35x 4x2 – 35x – 9 = 0 x = 9 or – (b) Solve: A 4(x + 4) > 9(x – 2) 4x + 16 > 9x – 18 16 + 18 > 9x – 4x 34 > 5x x < (c) (or 34 5 ) A Katya buys fish and chips for her family. She buys four pieces of fish and three scoops of chips. She spends $altogether. Each scoop of chips costs 60 cents more than a piece of fish. Calculate the price of piece of fish. 4f + 3s = 4f + 3(f + ) = A 7f + = f = A piece of fish is$ M No marks if solved by guess and check without equations! (d) Where does the graph of y = 12x2 – 17x + 6 cross the x-axis? When 12x2 – 17x + 6 = 0 A x = or M also award A if factorised y = (3x –2)(4x – 3) as a partial solution 3 Assessor’s use only (e) Find k so that x2 + k x + 16 has only one root. Only one root when discriminant Δ = 0 b2 – 4ac = 0 (f) k2 – 4 × 1 × 16 = 0 A k2 = 64 M k = ±8 E Find k so that 2x2 + k x + 50 has one root four times as large as the other. 2(x + r)(x + 4r) = 2x2 + k x + 50 A 2x2 + 10r x + 8r2 = 2x2 + k x + 50 So 8r2 = 50 which means r = ± so k = 10r k = ± 25 M E or 2r2 + k r + 50 = 0 and so 2r2 + k r + 50 = 0 2(4r)2 + k (4r) + 50 = 0 32r2 + 4k r + 50 = 0 and so 8r2 + 4k r + = 0 and 32r2 + 4k r + 50 = 0 r = ± which we put back into 2r2 + k r + 50 = 0 to get M k = ±25 E or roots are given by −𝑏±√𝑏2 −4𝑎𝑐 2𝑎 Designate them as r = So as 4r = 4( −𝑘−√𝑘 2 −4×2×50 2×2 −𝑘−√𝑘 2 −4×2×50 2×2 )= and 4r = −𝑘+√𝑘 2 −4×2×50 2×2 −𝑘+√𝑘 2 −4×2×50 A M 2×2 –4k – 4√𝑘 2 − = –k + √𝑘 2 − 5√𝑘 2 − = –3k squaring both sides 25(k2 – ) = 9k2 k = ± 25 16k2 – 10, = 0 E 4 Assessor’s use only QUESTION THREE (a) Write as the logb of a single number: 2 logb 6 – logb 4 36 2 log 6 – log 4 = log 62 – log 4 = log ( 4 ) = log 9 (b) Solve: must be fully simplified A logx = 3 log x = 3 so x3 = 3 x = √ = 5 (c) Solve: must have some working, not just the answer A 34 = 4x 34 = 4x log(34) = log(4x) = x log(4) x = log(34) ÷ log(4) x = A must have at least one stage (any of above) of working, not just the answer (d) A fully charged car battery has amps. However, the lights have been left on and so the end of each hour, the amperage of the battery is only 95% of what it was at the end of the previous hour. The amperage of the battery is given by the formula; Amps = × t where t = time in hours The car will not start if the amperage remaining is less than amps. Calculate how long before the car will not start = × t = t log() = log(t) = t log() A t = log() ÷ log() It takes hours (= 27 hours 2 minutes) must have at least one stage (any of above) of working, not just the answer. 5 M Assessor’s use only (e) If 16x + 1 + 16x = 68, find 22x 16x × 16 + 16x × 1 = 68 16x × 17 = 68 16x = 4 M 42x = 4 2x = 1 22x = 21 = 2 E (x = ½ but you don’t need to solve to answer the question) (f) Solve: 2k = 13 + 𝟕 𝒌 The easiest way is to multiply through by k. so 2k2 – 13k – 7 = 0 2k2 = 13k + 7 k = 7 or – E or 2k = 13𝑘 7 + 𝑘 𝑘 2k2 = 13k + 7 so 2k = + 13𝑘 + 7 𝑘 etc For each question: N0: No relevant evidence N1: Attempt at one question N2: 1 A A3: 2 A A4: 3A or 1M with 1 A M or 1 M M5: 1 M M6: 2 M E7: 1E E8: 2E Overall grade 0–8 = Not Achieved, 9–14 = Achieved, 15–19 = Merit, 20+ = Excellence 6

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## Simplify (3x^2 - 15x)/(3x^2 - 13x )

Begin by considering the numerator and denominator separately, simplify both:

1) 3x^2 - 15x

Here, a common factor of 3x can be removed to give 3x(x^2 -5).

2) 3x^2 - 13x

This involves using two brackets each arranged as (ax+b)(cx+d). Unlike in 1), 3x cannot be removed as a common factor, but 3x must exist in order for 3x^2 to be present, therefore you get (3x+b)(x+d). The next thing to consider is how to multiply b and d to obtain 10, the factors of 10 are 1, 10, 2 and 5. As "" is in the expression to be simplified, one of the signs must be a minus. So considering 2 and 5, there are two possible answers: (3x - 2)(x+5) or (3x+2)(x-5), multiplying these out you find that the latter is the only possible solution.

Finally, putting the new simplified fraction together you get: (3x(x-5))/((3x+2)(x-5), (x-5) cancel on the numerator and denominator giving you a final, simplified, fraction of 3x/(3x+2).  ### Most Used Actions

 \mathrm{simplify} \mathrm{solve\:for} \mathrm{expand} \mathrm{factor} \mathrm{rationalize}
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