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f(x) = 3x^3 - 5x^2 - 58x + 40 (a) Find the remainder when f(x) is divided by (x - 3) - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 3

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f(x)-=-3x^3---5x^2---58x-+-40--(a)-Find-the-remainder-when-f(x)-is-divided-by-(x---3)-Edexcel-A-Level Maths Pure-Question 4-2010-Paper 3.png

f(x) = 3x^3 - 5x^2 - 58x + 40 (a) Find the remainder when f(x) is divided by (x - 3). Given that (x - 5) is a factor of f(x), (b) find all the solutions of f(x) = ... show full transcript

Worked Solution & Example Answer:f(x) = 3x^3 - 5x^2 - 58x + 40 (a) Find the remainder when f(x) is divided by (x - 3) - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 3

Step 1

Find the remainder when f(x) is divided by (x - 3)

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Answer

To find the remainder when dividing a polynomial by a linear factor, we can use the Remainder Theorem. This states that the remainder of the division of a polynomial f(x) by (x - c) is equivalent to f(c). Here, we calculate f(3):

f(3)=3(3)35(3)258(3)+40=3(27)5(9)174+40=8145174+40=8145+40174=8145+40174=98. f(3) = 3(3)^3 - 5(3)^2 - 58(3) + 40 = 3(27) - 5(9) - 174 + 40 = 81 - 45 - 174 + 40 = 81 - 45 + 40 - 174 = 81 - 45 + 40 - 174 = -98.

Thus, the remainder when f(x) is divided by (x - 3) is -98.

Step 2

Find all the solutions of f(x) = 0

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Answer

Since (x - 5) is given as a factor of f(x), we can factor f(x) as follows:

Let f(x) = (x - 5)(Ax^2 + Bx + C) for some constants A, B, and C. Using polynomial long division on f(x):

  1. Divide f(x) by (x - 5) to find the quadratic.

Perform the polynomial long division, which yields:

f(x)=(x5)(3x2+10x8). f(x) = (x - 5)(3x^2 + 10x - 8).
  1. Now, set the factored form equal to zero:
(x5)(3x2+10x8)=0.(x - 5)(3x^2 + 10x - 8) = 0.

This gives us the first solution: x = 5.

  1. Next, solve the quadratic equation 3x2+10x8=03x^2 + 10x - 8 = 0 using the quadratic formula:
x=b±b24ac2a, x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},

where a = 3, b = 10, and c = -8:

x=10±1024(3)(8)2(3)=10±100+966=10±1966=10±146. x = \frac{-10 \pm \sqrt{10^2 - 4(3)(-8)}}{2(3)} = \frac{-10 \pm \sqrt{100 + 96}}{6} = \frac{-10 \pm \sqrt{196}}{6} = \frac{-10 \pm 14}{6}.
  1. This gives us the solutions:
x1=46=23 x_1 = \frac{4}{6} = \frac{2}{3}

and

x2=246=4. x_2 = \frac{-24}{6} = -4.

Thus, the complete set of solutions to f(x) = 0 is x = 5, x = \frac{2}{3}, and x = -4.

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