2016 APE Tutoring Junior High School Math contest training Camp homework answer-5

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1. Decomposition $a ^3-4a^2 + A + 6$.

Answer:

Make $f (a) = a^3-4a^2 + A + 6$, its rational root may be $\pm1$, $\pm2$, $\pm3$, $\pm6$.

Note that $f (a) $ odd term and even secondary coefficients and equal (equal to $2$), so $f (-1) = 0$. By comprehensive division can be: $ $a ^3-4a^2 + a-6 = (A + 1) (a^2-5a + 6) = (A + 1) (a-2) (a-3). $$

2. Decomposition of the $a ^4 + 3a^3-3a^2-11a-6$.

Answer:

Make $f (a) = A^4 + 3a^3-3a^2-11a-6$, its rational root may be $\pm1$, $\pm2$, $\pm3$, $\pm6$.

$f (a) $ odd term and even secondary coefficients and equal (equals $-8$), so $f (-1) = 0$. By comprehensive division You can get: $ $a ^4 + 3a^3-3a^2-11a-6 = (A + 1) (a^3 + 2a^2-5a-6). $$ Notice that the second parenthesis has the odd and even second coefficients and is equal (equal to $-4$), so $a +1$ is its $ $a ^4 + 3a^3-3a^2-11a-6 = (A + 1) (a^3 + 2a^2-5a-6) $$ $$= (A + 1) (A + 1) (a^2 + a-6) = (a+1) ^2 (A + 3) (a-2). $$

3. Decomposition of the $x ^4-x^3y-7x^2y^2 + 13xy^3-6y^4$.

Answer:

Consider $x $ as a principal, even if $f (x) = x^4-x^3y-7x^2y^2 + 13xy^3-6y^4$, its rational root may be $\pm y$, $\pm 2y$, $\pm 3y$, $\pm 6y$, $\pm y^2$, $\ PM 2y^2$, $\cdots$, $\pm 6y^4$.

Verify $f (y) = y^4-y^4-7y^4 + 13y^4-6y^4 = 0$, so the combined division can be $ $x ^4-x^3y-7x^2y^2 + 13xy^3-6y^4 = (x-y) (x^3-7xy^2 + 6y^3). $$ The second parenthesis continues with the $x $ as the main element $x-y$ is also its cause, that is, $ $x ^4-x^3y-7x^2y^2 + 13xy^3-6y^4 = (x-y) (x^3-7xy^2 + 6y^3) $$ $$= ( XY) ^2 (x^2 + xy-6y^2) = (x-y) ^2 (x-2y) (× + 3y). $$

4. Known polynomial $f (x) = x^5 + 3x^4 + 8x^3+ 11x + k$ can be divisible by $x +2$, $k $ value.

Answer:

A $ $f ( -2) = -32 + 48-64-22 + k = 0 \rightarrow k = 70.$$ by the formula theorem

5. Set $f (x) = x^2 + mx + n$ ($m, n$ are integers) both as polynomial $x ^4 + 6x^2 + 25$, and as a polynomial $3x^4 + 4x^2 + 28x + 5$ for $f (x) $ and $f (-1) $ .

Answer:

The difference method is considered to solve the problem of dividing the polynomial evenly.

$ $f (x) \ \big{|} \ \left (x^4 + 6x^2 + 25\right) \rightarrow f (x) \ \big{|} \ 3\left (x^4 + 6x^2 + 25\right) = 3x^4 + 18x^2 + 75,$$ $$\rightarrow f (x) \ \big{|} \ \left (3x^4 + 18x^2 + 75\right)-\left (3x^4 + 4x^2 + 28x + 5\right) $$ $$= 14x^2-28x + = 14\left (x^2-2x + 5\right) , $$ $$\rightarrow f (x) = x^2 + mx + n = x^2-2x + 5 \rightarrow F (-1) = 8.$$

6. Verification: $a-b$, $b-c$, $c-a$ are $a ^2 (b-c) + b^2 (c-a) + c^2 (A-B) $ of the type, and decomposition of the formula.

Answer:

The $a $ is treated as a principal, even if $f (a) = A^2 (b-c) + b^2 (c-a) + c^2 (A-B). $

Easy to get $f (b) = b^2 (b-c) + b^2 (c-b) + c^2 (b-b) = 0$, i.e. $a-b$ is $f (a) $.

Similarly, $b-c$, $c-a$ are their own.

Because the original is three times homogeneous polynomial, its three factors are one-time, so set $a ^2 (b-c) + b^2 (c-a) + c^2 (A-B) = K (a) (B-C) (c-a) $ (where $k $ is the undetermined factor),

Make $a = 0$, $b = 1$, $c = 2$ into $k = -1$. So $ $a ^2 (b-c) + b^2 (c-a) + c^2 (A-B) =-(A-B) (b-c) (c-a). $$

2016 APE Tutoring Junior High School Math contest training Camp homework answer-5