How do you factor the trinomial $56 r^{2} + 121 r + 63$ $56r^2+121r+63$ ?

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How do you factor the trinomial $56 r^{2} + 121 r + 63$ $56r^2+121r+63$ ?

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Answer 1 · 2015-12-17T21:49:18

Use the quadratic formula to find zeros, hence factors:

$56 r^{2} + 121 r + 63 = (7 r + 9) (8 r + 7)$ $56r^2+121r+63 = (7r+9)(8r+7)$

Explanation:

$f (r) = 56 r^{2} + 121 r + 63$ $f(r) = 56r^2+121r+63$ is of the form $a r^{2} + b r + c$ $ar^2+br+c$ with $a = 56$ $a=56$ , $b = 121$ $b=121$ and $c = 63$ $c=63$ .

This has zeros for values of $r$ $r$ given by the quadratic formula:

$r = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $r = (-b+-sqrt(b^2-4ac))/(2a)$

$= \frac{- 121 \pm \sqrt{121^{2} - (4 \times 56 \times 63)}}{2 \times 56}$ $=(-121+-sqrt(121^2-(4xx56xx63)))/(2xx56)$

$= \frac{- 121 \pm \sqrt{14641 - 14112}}{112}$ $=(-121+-sqrt(14641-14112))/112$

$= \frac{- 121 \pm \sqrt{529}}{112}$ $=(-121+-sqrt(529))/112$

$= \frac{- 121 \pm 23}{112}$ $=(-121+-23)/112$

That is $r = - \frac{144}{112} = - \frac{9}{7}$ $r = -144/112 = -9/7$ or $r = - \frac{98}{112} = - \frac{7}{8}$ $r = -98/112 = -7/8$

Hence:

$56 r^{2} + 121 r + 63 = (7 r + 9) (8 r + 7)$ $56r^2+121r+63 = (7r+9)(8r+7)$

Answer 2 · 2015-12-17T22:18:08

Alternatively use prime factorisation and reasoning to find:

$56 r^{2} + 121 r + 63 = (7 r + 9) (8 r + 7)$ $56r^2+121r+63 = (7r+9)(8r+7)$

Explanation:

Alternatively, reason your way to the integer factorisation as follows:

$56 r^{2} + 121 r + 63$ $56r^2+121r+63$

The coefficients have the following prime factorisations:

$56 = 2 \cdot 2 \cdot 2 \cdot 7$ $56 = 2*2*2*7$

$121 = 11 \cdot 11$ $121= 11*11$

$63 = 3 \cdot 3 \cdot 7$ $63 = 3*3*7$

Notice that both the leading and trailing terms are divisible by $7$ $7$ but the middle term is not. That means that if there is an exact factorisation with integer coefficients then it can take the form:

$56 r^{2} + 121 r + 63 = (7 a r + b) (c r + 7 d)$ $56r^2+121r+63 = (7ar+b)(cr+7d)$

$= 7 a c r^{2} + (49 a d + b c) r + 7 b d$ $= 7acr^2+(49ad+bc)r+7bd$

for some integers $a$ $a$ , $b$ $b$ , $c$ $c$ and $d$ $d$ , where we can take $a > 0$ $a > 0$ without loss of generality.

Notice also that the middle term is odd, so one of $a$ $a$ and $c$ $c$ is odd. Hence either $a = 8$ $a=8$ and $c = 1$ $c=1$ or $a = 1$ $a=1$ and $c = 8$ $c=8$ .

Similarly, notice that the middle term is not divisible by $3$ $3$ , hence either $b = 9$ $b=9$ and $d = 1$ $d=1$ or $b = 1$ $b=1$ and $d = 9$ $d=9$ . $b$ $b$ and $d$ $d$ cannot be negative since the constant term is positive and the middle term is also positive.

This gives us $4$ $4$ possibilities to try to match the middle term:

$49 a d + b c = 121$ $49ad+bc = 121$

with $(a, b, c, d) = (8, 1, 9, 1), (1, 8, 9, 1), (8, 1, 1, 9) or (1, 8, 1, 9)$ $(a, b, c, d) = (8, 1, 9, 1), (1, 8, 9, 1), (8, 1, 1, 9) or (1, 8, 1, 9)$

We quickly find $a = 1$ $a=1$ , $c = 8$ $c=8$ , $d = 1$ $d=1$ and $b = 9$ $b=9$

Hence:

$56 r^{2} + 121 r + 63 = (7 r + 9) (8 r + 7)$ $56r^2+121r+63 = (7r+9)(8r+7)$

Answer 3 · 2015-12-17T22:45:43

Alternatively use an AC Method to find:

$56 r^{2} + 121 r + 63 = (7 r + 9) (8 r + 7)$ $56r^2+121r+63=(7r+9)(8r+7)$

Explanation:

Given $56 r^{2} + 121 r + 63$ $56r^2+121r+63$ look for a pair of factors of $A C = 56 \cdot 63 = 3528$ $AC = 56*63 = 3528$ whose sum is $B = 121$ $B=121$ .

First note that $56 + 63 = 119$ $56+63 = 119$ is very close and to get a greater sum, the pair of factors needs to be slightly further apart.

The prime factorisation of $3528$ $3528$ is:

$3528 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 7 \cdot 7$ $3528 = 2*2*2*3*3*7*7$

Also note that $121 = 11 \cdot 11$ $121=11*11$ is not divisible by $2$ $2$ , $3$ $3$ or $7$ $7$ , so all of the $2$ $2$ 's must be in one of the factors, all of the $3$ $3$ 's in one of the factors and all of the $7$ $7$ 's in one of the factors.

We quickly find $49 \times 72 = 3528$ $49xx72 = 3528$ with $49 + 72 = 121$ $49+72=121$

Hence we can factor by grouping:

$56 r^{2} + 121 r + 63$ $56r^2+121r+63$

$= (56 r^{2} + 49 r) + (72 r + 63)$ $=(56r^2+49r)+(72r+63)$

$= 7 r (8 r + 7) + 9 (8 r + 7)$ $=7r(8r+7)+9(8r+7)$

$= (7 r + 9) (8 r + 7)$ $=(7r+9)(8r+7)$

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